Suppose that $\:|\psi\rangle \in \mathrm F\:$, where $\:\mathrm F\:$ a $\:m$-dimensional
Hilbert space with basic state vectors $\:|f_{i}\rangle, \quad i=1,2\cdots,m\:$ and $\:|\phi\rangle \in \mathrm H\:$, where $\:\mathrm H\:$ a $\:n$-dimensional
Hilbert space with basic state vectors $\:|h_{j}\rangle, \quad j=1,2\cdots,n$.
To prove the identity
\begin{equation}
\left(|\psi\rangle \otimes |\phi\rangle\right)\left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger} = \left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right)
\tag{01}
\end{equation}
it's sufficient to prove the identity for the basic state vectors
\begin{equation}
\left(|f_{i}\rangle \otimes |h_{j}\right)\left(|f_{k}\rangle \otimes |h_{\ell}\right)^{\dagger} = \left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)
\tag{02}
\end{equation}
since if (using Einstein's summation convention)
\begin{align}
|\psi\rangle & = a_{i}|f_{i}\rangle \qquad i=1,2\cdots,m
\tag{03a}\\
|\phi\rangle & = b_{j}|h_{j}\rangle \qquad j=1,2\cdots,n
\tag{03b}
\end{align}
then
\begin{align}
\left(|\psi\rangle \otimes |\phi\rangle\right) & = a_{i}b_{j}\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)
\tag{04a}\\
\left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger} & = a^*_{k}b^*_{\ell}\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\dagger}
\tag{04b}
\end{align}
and
\begin{align}
\left(|\psi\rangle \otimes |\phi\rangle\right)\left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger} & = a_{i}a^*_{k}b_{j}b^*_{\ell}\left(|f_{i}\rangle \otimes |h_{j}\right)\left(|f_{k}\rangle \otimes |h_{\ell}\right)^{\dagger}
\tag{05a}\\
\left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right) & = a_{i}a^*_{k}b_{j}b^*_{\ell}\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)
\tag{05b}
\end{align}
Note that
\begin{align}
& \left(|\psi\rangle \otimes |\phi\rangle\right)\left(|\psi\rangle \otimes |\phi\rangle\right)^{\dagger}
\tag{06a}\\
& \left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right)
\tag{06b}\\
& \left(|f_{i}\rangle \otimes |h_{j}\rangle\right)\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\dagger}
\tag{06c}\\
& \left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)
\tag{06d}
\end{align}
are all linear transformations in the product $\:(m\!\cdot\! n)$-dimensional space $\: \mathrm F \otimes \mathrm H$.
To prove identity (02) we may take without loss of generality as basic state vectors those having a component equal to 1 and all other components equal to 0
\begin{align}
\left(|f_{i}\rangle \right)_{\mu} & = \delta_{i\mu} , \quad \left(|f_{k}\rangle \right)_{\rho} = \delta_{k\rho} \qquad i,k,\mu,\rho =1,2,\cdots, m
\tag{07a}\\
\left(|h_{j}\rangle \right)_{\nu} & = \delta_{j\nu} , \quad \left(|h_{\ell}\rangle \right)_{\sigma} = \delta_{\ell\sigma} \qquad j,\ell,\nu,\sigma =1,2,\cdots, n
\tag{07b}
\end{align}
Then the transformations (06c) and (06d) are equal and are represented by a $\:(m\!\cdot\! n)\times(m\!\cdot\! n)$ matrix with one element equal to 1 and all other elements equal to 0.
By example : Let $\:m=3,n=2\:$ and $\:i=3,k=2, j=2, \ell=1\:$. Then
\begin{equation}
\left(|f_{i}\rangle \otimes |h_{j}\rangle\right) =\left(|f_{3}\rangle \otimes |h_{2}\rangle\right) =
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\otimes
\begin{bmatrix}
0\\
1
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
1
\end{bmatrix}
\tag{08}
\end{equation}
\begin{equation}
\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right) =\left(|f_{2}\rangle \otimes |h_{1}\rangle\right) =
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
\otimes
\begin{bmatrix}
1\\
0
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
1\\
0\\
0\\
0
\end{bmatrix}
\tag{09}
\end{equation}
so
\begin{equation}
\left(|f_{3}\rangle \otimes |h_{2}\rangle\right)\left(|f_{2}\rangle \otimes |h_{1}\rangle\right)^{\dagger} =
\begin{bmatrix}
0\\
0\\
0\\
0\\
0\\
1
\end{bmatrix}
\cdot
\begin{bmatrix}
0\\
0\\
1\\
0\\
0\\
0
\end{bmatrix}^{\boldsymbol{\dagger}}
=
\begin{bmatrix}
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&1&0&0&0\\
\end{bmatrix}
\tag{10}
\end{equation}
Now
\begin{equation}
\left(|f_{i}\rangle\langle f_{k}|\right) = \left(|f_{3}\rangle\langle f_{2}|\right)=
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\begin{bmatrix}
0&1&0
\end{bmatrix}
=
\begin{bmatrix}
0&0&0\\
0&0&0\\
0&1&0
\end{bmatrix}
\tag{11}
\end{equation}
and
\begin{equation}
\left(|h_{j}\rangle\langle h_{\ell}|\right) = \left(|h_{1}\rangle\langle h_{2}|\right)=
\begin{bmatrix}
0\\
1
\end{bmatrix}
\begin{bmatrix}
1&0
\end{bmatrix}
=
\begin{bmatrix}
0&0\\
1&0
\end{bmatrix}
\tag{12}
\end{equation}
so
\begin{equation}
\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right) = \left(|f_{3}\rangle\langle f_{2}|\right)\otimes \left(|h_{1}\rangle\langle h_{2}|\right) =
\begin{bmatrix}
0&0&0\\
0&0&0\\
0&1&0
\end{bmatrix}
\otimes
\begin{bmatrix}
0&0\\
1&0
\end{bmatrix}
\tag{13}
\end{equation}
that is
\begin{equation}
\left(|f_{3}\rangle\langle f_{2}|\right)\otimes \left(|h_{1}\rangle\langle h_{2}|\right) =
\begin{bmatrix}
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&1&0&0&0\\
\end{bmatrix}
\tag{14}
\end{equation}
From ((10) and (14)
\begin{equation}
\left(|f_{3}\rangle \otimes |h_{2}\rangle\right)\left(|f_{2}\rangle \otimes |h_{1}\rangle\right)^{\dagger}
=
\begin{bmatrix}
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&1&0&0&0\\
\end{bmatrix}
=
\left(|f_{3}\rangle\langle f_{2}|\right)\otimes \left(|h_{1}\rangle\langle h_{2}|\right)
\tag{15}
\end{equation}
Note :
I would prefer the following symbols that give an elegant expression of identity (01). More exactly, since $\:\left(|\psi\rangle \otimes |\phi\rangle\right)\:$ is a state in the product space $\: \mathrm F \otimes \mathrm H\:$ we define it as a ket
\begin{equation}
|\psi\otimes \phi\rangle \stackrel{\text{def}}{\equiv} \left(|\psi\rangle \otimes |\phi\rangle\right)
\tag{N-01}
\end{equation}
Then
\begin{equation}
\left(|\psi\rangle \otimes |\phi\rangle\right)^{\boldsymbol{\dagger}}=
\langle \psi\otimes \phi|
\tag{N-02}
\end{equation}
and identity (01) is expressed as
\begin{equation}
\color{blue}{|\psi\otimes \phi\rangle\langle \psi\otimes \phi| = \left(|\psi\rangle\langle\psi|\right)\otimes \left(|\phi\rangle\langle\phi|\right)}
\color{black}{}
\tag{N-03}
\end{equation}
Now, if each of $\:|\psi\rangle \in \mathrm F,|\phi\rangle \in \mathrm H\:$ is a unit ket in its own Hilbert space
\begin{equation}
\langle\psi|\psi\rangle = 1 = \langle\phi|\phi\rangle
\tag{N-04}
\end{equation}
then in $\: \mathrm F \otimes \mathrm H\:$
\begin{equation}
\langle \psi\otimes \phi|\psi\otimes \phi\rangle = 1
\tag{N-05}
\end{equation}
In this case we have the following interpretation of (N-03) :
The projection on the unit ket $\:|\psi\otimes \phi\rangle\:$ in the product space $\: \mathrm F \otimes \mathrm H\:$ is the product of the projection on the unit ket $\:|\psi\rangle\:$ in $\:\mathrm F\:$ by the projection on the unit ket $\:|\phi\rangle\:$ in $\:\mathrm H$.
From this example above, we could give a proof of identity (02) in general under the choice (07) for the basic state vectors. So
1. $\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)$ is a $(m\!\cdot\! n)$-column vector with its $(i\!-\!1)n\!+\!j$ component equal to 1 and all the other components equal to 0
\begin{equation}
\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)=
\begin{bmatrix}
0\\
0\\
\vdots\\
0\\
\vdots\\
\color{blue}{\bf 1}\\
\vdots\\
0
\end{bmatrix}
\begin{matrix}
\longleftarrow \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow
\end{matrix}
\begin{matrix}
1\hphantom{(i\!-\!1)n\!+\!j}\\
2\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j} \\
\kappa\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\color{red}{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\:m\cdot n\hphantom{(i\!-\!1)n}
\end{matrix}
\tag{16}
\end{equation}
2. $\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}}\:$ is a $(m\!\cdot\! n)$-row vector with its $(k\!-\!1)n\!+\!\ell$ component equal to 1 and all the other components equal to 0
\begin{align}
\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}} & =
\begin{bmatrix}
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}\color{blue}{\bf 1}\hphantom{)n\!+\!\ell}&\cdots&0\\
\end{bmatrix}
\tag{17}\\
& \hphantom{=\!=}
\begin{matrix}
\:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\hphantom{k\!-\!1}\uparrow\hphantom{n\!+\!\ell}&\cdots&\uparrow
\end{matrix}
\nonumber\\
& \hphantom{=\!=}
\begin{matrix}
\:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{(k\!-\!1)n\!+\!\ell}&\,\cdots&\!\!m\cdot n
\end{matrix}
\nonumber
\end{align}
3. $\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}}$, product of equations (16) and(17), is a $(m\!\cdot\! n)\times(m\!\cdot\! n)$-square matrix with its element in the $(i\!-\!1)n\!+\!j$ row and $(k\!-\!1)n\!+\!\ell$ column equal to 1 and all the other elements equal to 0
\begin{align}
\left(|f_{i}\rangle \otimes |h_{j}\rangle\right)&\left(|f_{k}\rangle \otimes |h_{\ell}\rangle\right)^{\boldsymbol{\dagger}} =
\tag{18}\\
&
\begin{bmatrix}
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}\color{blue}{\bf 1}\hphantom{)n\!+\!\ell}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0
\end{bmatrix}
\begin{matrix}
\longleftarrow \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow
\end{matrix}
\begin{matrix}
1\hphantom{(i\!-\!1)n\!+\!j}\\
2\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j} \\
\kappa\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\color{red}{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\:m\cdot n\hphantom{(i\!-\!1)n}
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\hphantom{k\!-\!1}\uparrow\hphantom{n\!+\!\ell}&\cdots&\uparrow
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{(k\!-\!1)n\!+\!\ell}&\,\cdots&\!\!m\cdot n
\end{matrix}
\nonumber
\end{align}
4. $\left(|f_{i}\rangle\langle f_{k}|\right)$ is a $m\!\times\! m$-square matrix with its element in the $i$-row and $k$-column equal to 1 and all the other elements equal to 0
\begin{align}
\left(|f_{i}\rangle\langle f_{k}|\right) & =
\tag{19}\\
&
\begin{bmatrix}
0&0&\cdots&0&\cdots&0&\cdots&0\\
0&0&\cdots&0&\cdots&0&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
0&0&\cdots&0&\cdots&0&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\color{blue}{\bf 1}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
0&0&\cdots&0&\cdots&0&\cdots&0
\end{bmatrix}
\begin{matrix}
\longleftarrow \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow
\end{matrix}
\begin{matrix}
1\hphantom{(i\!-\!1)n\!+\!j}\\
2\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j} \\
\kappa\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\color{red}{i}\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\:m\hphantom{(i\!-\!1)n\!+\!j}
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\uparrow&\cdots&\uparrow
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{k}&\,\cdots&\!\!m
\end{matrix}
\nonumber
\end{align}
5. $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ is a $n\!\times\!n$-square matrix with its element in the $j$-row and $\ell$-column equal to 1 and all the other elements equal to 0
\begin{align}
\left(|h_{j}\rangle\langle h_{\ell}|\right) & =
\tag{20}\\
&
\begin{bmatrix}
0&0&\cdots&0&\cdots&0&\cdots&0\\
0&0&\cdots&0&\cdots&0&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
0&0&\cdots&0&\cdots&0&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\color{blue}{\bf 1}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
0&0&\cdots&0&\cdots&0&\cdots&0
\end{bmatrix}
\begin{matrix}
\longleftarrow \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow
\end{matrix}
\begin{matrix}
1\hphantom{(i\!-\!1)n\!+\!j}\\
2\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j} \\
\kappa\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\color{red}{j}\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\:n\hphantom{(i\!-\!1)n\!+\!j}
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\uparrow&\cdots&\uparrow
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{\ell}&\,\cdots&\!\!n
\end{matrix}
\nonumber
\end{align}
6. $\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)$ is a $m\!\times\!m$-square matrix in block form. All its block elements are $n\!\times\!n$-square null matrices (symbol $\mathcal{O}_{n}$) except the block in the $i$-row and $k$-column which is equal to the $n\!\times\!n$-square matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$. Of course $\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)$ is a $(m\!\cdot\! n)\times(m\!\cdot\! n)$-square matrix. In block form
\begin{align}
\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right) & =
\tag{21}\\
&
\begin{bmatrix} \mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\
\mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
\mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
\mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots& \color{blue}{\left(|h_{j}\rangle\langle h_{\ell}|\right)}&\cdots&\mathcal{O}_{n}\\
\vdots&\vdots&\cdots&\vdots&\cdots&\vdots&\cdots&\vdots\\
\mathcal{O}_{n}&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}&\cdots&\mathcal{O}_{n}\\
\end{bmatrix}
\begin{matrix}
\longleftarrow \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow
\end{matrix}
\begin{matrix}
1\hphantom{(i\!-\!1)n\!+\!j}\\
2\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j} \\
\kappa\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\color{red}{i}\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\:m\hphantom{(i\!-\!1)n\!+\!j}
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:\:\,\uparrow & \:\:\:\,\uparrow &\,\,\cdots &\:\,\uparrow &\,\,\cdots &\hphantom{==}\uparrow &\hphantom{==}\cdots &\:\:\uparrow
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:\:\,1 & \:\:\:\,2 & \,\,\cdots &\:\, \lambda &\,\,\cdots &\hphantom{==}\,\color{red}{k} &\hphantom{==}\cdots &\:\: m
\end{matrix}
\nonumber
\end{align}
Now, in this block form we note that above the matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ there are $(i\!-\!1)$ rows of $n\!\times\!n$-square null matrices and to the left of the matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ there are $(k\!-\!1)$ columns of $n\!\times\!n$-square null matrices. This means that the matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ is located after the element in the $(i-1)n$ row and the $(k-1)n$ column. But inside this matrix $\left(|h_{j}\rangle\langle h_{\ell}|\right)$ the unique nonzero element (=1) is located in the $j$-row and $\ell$-column and so in the $(i\!-\!1)n\!+\!j$ row and $(k\!-\!1)n\!+\!\ell$ column of the matrix $\left(|f_{i}\rangle\langle f_{k}|\right)\otimes \left(|h_{j}\rangle\langle h_{\ell}|\right)$. So
\begin{align}
\left(|f_{i}\rangle\langle f_{k}|\right)\otimes & \left(|h_{j}\rangle\langle h_{\ell}|\right) =
\tag{22}\\
&
\begin{bmatrix}
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}\color{blue}{\bf 1}\hphantom{)n\!+\!\ell}&\cdots&0\\
\vdots&\vdots&\cdots&\vdots&\cdots&\hphantom{(k\!-\!1}\vdots\hphantom{)n\!+\!\ell}&\cdots&\vdots\\
0&0&\cdots&0&\cdots&\hphantom{(k\!-\!1}0\hphantom{)n\!+\!\ell}&\cdots&0
\end{bmatrix}
\begin{matrix}
\longleftarrow \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow \\
\vdots \\
\longleftarrow
\end{matrix}
\begin{matrix}
1\hphantom{(i\!-\!1)n\!+\!j}\\
2\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j} \\
\kappa\hphantom{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\color{red}{(i\!-\!1)n\!+\!j}\\
\vdots\hphantom{(i\!-\!1)n\!+\!j}\\
\:m\cdot n\hphantom{(i\!-\!1)n}
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:\uparrow &\uparrow &\cdots&\uparrow&\cdots&\hphantom{k\!-\!1}\uparrow\hphantom{n\!+\!\ell}&\cdots&\uparrow
\end{matrix}
\nonumber\\
& \hphantom{.\,}
\begin{matrix}
\:\:1 & 2 & \cdots & \lambda &\cdots&\color{red}{(k\!-\!1)n\!+\!\ell}&\,\cdots&\!\!m\cdot n
\end{matrix}
\nonumber
\end{align}
Equations (18) and (22) give a proof of the identity (02) and so a proof of the identity (01).
