Passing from curl to vector product I don't understand how to obtain second equation with first part in the equation
$$
\nabla \times \vec A_0 e^{-j \vec k\cdot \vec r} = -j\vec k\times \vec A_0 e^{-j \vec k\cdot \vec r}.
$$
Can you show me how to derive it?
 A: The $i$th component of the left-hand side is $\epsilon_{ijk}\partial_j(A_{0k}e^{-j\vec{k}\cdot\vec{r}})$, where $\sum_{jk}$ is implicit. (Don't confuse dummy indices with the square root of $-1$.) The $i$th component of the right-hand side is $-j\epsilon_{ijk}k_jA_{0k}e^{-j\vec{k}\cdot\vec{r}}$. The rest is the product rule.
A: \begin{equation}
\boldsymbol{\nabla}\equiv
\begin{bmatrix}
\dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}\,, \quad
\mathbf{A}=
\begin{bmatrix}
A_{1}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{2}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\end{bmatrix}\,, \quad
\mathbf{k}\boldsymbol{\cdot}\mathbf{x}=k_{1}x_{1}+k_{2}x_{2}+k_{3}x_{3}
\tag{01}
\end{equation}
\begin{align}
\boldsymbol{\nabla}\boldsymbol{\times}\left(\mathbf{A}\,e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\right) & =
\begin{bmatrix}
\mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}} &  \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{1}e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} & A_{2}e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} & A_{3}e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}} \vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
=
\begin{bmatrix}
A_{3}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{2}}-A_{2}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{1}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{3}}-A_{3}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{2}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{1}}-A_{1}\dfrac{\partial e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
\nonumber\\
& =\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}
\begin{bmatrix}
k_{2}A_{3}-k_{3}A_{2}\vphantom{\dfrac{\dfrac{}{}}{}}\\
k_{3}A_{1}-k_{1}A_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
k_{1}A_{2}-k_{2}A_{1}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
=\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}
\underbrace{
\begin{bmatrix}
\mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
k_{1} & k_{2} &  k_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{1} & A_{2} &  A_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}}_{\mathbf{k}\boldsymbol{\times}\mathbf{A}}
\tag{02}
\end{align}
so
\begin{equation}
\boxed{\:\boldsymbol{\nabla}\boldsymbol{\times}\left(\mathbf{A}\,e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\right) =\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\left(\mathbf{k}\boldsymbol{\times}\mathbf{A}\right)\:\vphantom{\dfrac12^{\tfrac12}_{\tfrac12}}}
\tag{03}
\end{equation}

More generally : 
If $\:\psi\left(x_{1},x_{2},x_{3}\right)\:$ and $\:\mathbf{A}\left(x_{1},x_{2},x_{3}\right)\:$
are scalar and vector functions respectively of the coordinates in $\:\mathbb{R}^{3}\:$ then
\begin{equation}
\boxed{\:\boldsymbol{\nabla}\boldsymbol{\times}\left(\psi\mathbf{A}\right)=\boldsymbol{\nabla}\psi\boldsymbol{\times}\mathbf{A}+\psi\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}\:\vphantom{\dfrac12^{\tfrac12}_{\tfrac12}}}
\tag{04}
\end{equation}
Equation (03) is a special case of (04) with 
\begin{equation}
\psi\left(\mathbf{x}\right)\equiv e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\,,\quad \mathbf{A}\left(\mathbf{x}\right)=\text{constant}
\tag{05}
\end{equation}
and the fact that 
\begin{equation}
\boldsymbol{\nabla}\psi=\boldsymbol{\nabla} e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}=\boldsymbol{-}\mathrm{i}\:e^{\boldsymbol{-}\mathrm{i}\:\mathbf{k}\boldsymbol{\cdot}\mathbf{x}}\mathbf{k}
\tag{06}
\end{equation}


Proof of identity (04):
\begin{align}
\boldsymbol{\nabla}\boldsymbol{\times}\left(\psi\mathbf{A}\right) & =
\begin{bmatrix}
\mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}} &  \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\psi A_{1} & \psi A_{2} & \psi A_{3} \vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
=
\begin{bmatrix}
\dfrac{\partial \left(\psi A_{3}\right)}{\partial x_{2}}-\dfrac{\partial \left(\psi A_{2}\right)}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial \left(\psi A_{1}\right)}{\partial x_{3}}-\dfrac{\partial \left(\psi A_{3}\right)}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial \left(\psi A_{2}\right)}{\partial x_{1}}-\dfrac{\partial \left(\psi A_{1}\right)}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
=
\begin{bmatrix}
A_{3}\dfrac{\partial \psi}{\partial x_{2}}+ \psi\dfrac{\partial A_{3}}{\partial x_{2}}-A_{2}\dfrac{\partial \psi}{\partial x_{3}}-\psi\dfrac{\partial A_{2}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{1}\dfrac{\partial \psi}{\partial x_{3}}+ \psi\dfrac{\partial A_{1}}{\partial x_{3}}-A_{3}\dfrac{\partial \psi}{\partial x_{1}}-\psi\dfrac{\partial A_{3}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{2}\dfrac{\partial \psi}{\partial x_{1}}+ \psi\dfrac{\partial A_{2}}{\partial x_{1}}-A_{1}\dfrac{\partial \psi}{\partial x_{2}}-\psi\dfrac{\partial A_{1}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
\nonumber\\
& =
\begin{bmatrix}
A_{3}\dfrac{\partial \psi}{\partial x_{2}}-A_{2}\dfrac{\partial \psi}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{1}\dfrac{\partial \psi}{\partial x_{3}}-A_{3}\dfrac{\partial \psi}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{2}\dfrac{\partial \psi}{\partial x_{1}}-A_{1}\dfrac{\partial \psi}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
+
\begin{bmatrix}
\psi\dfrac{\partial A_{3}}{\partial x_{2}}-\psi\dfrac{\partial A_{2}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\psi\dfrac{\partial A_{1}}{\partial x_{3}}-\psi\dfrac{\partial A_{3}}{\partial x_{1}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\psi\dfrac{\partial A_{2}}{\partial x_{1}}-\psi\dfrac{\partial A_{1}}{\partial x_{2}}\vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}
=
\underbrace{
\begin{bmatrix}
\mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial \psi}{\partial x_{1}} & \dfrac{\partial\psi}{\partial x_{2}} &  \dfrac{\partial\psi}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
 A_{1} &  A_{2} &  A_{3} \vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}}_{\boldsymbol{\nabla}\psi\boldsymbol{\times}\mathbf{A}}
+
\psi\,
\underbrace{
\begin{bmatrix}
\mathbf{e}_{1} & \mathbf{e}_{2} & \mathbf{e}_{3}\vphantom{\dfrac{\dfrac{}{}}{}}\\
\dfrac{\partial\hphantom{ x_{1}}}{\partial x_{1}} & \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{2}} &  \dfrac{\partial\hphantom{ x_{1}}}{\partial x_{3}}\vphantom{\dfrac{\dfrac{}{}}{}}\\
A_{1} & A_{2} &  A_{3} \vphantom{\dfrac{\dfrac{}{}}{}}
\end{bmatrix}}_{\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}}
\nonumber\\
& =\boldsymbol{\nabla}\psi\boldsymbol{\times}\mathbf{A}+\psi\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}
\tag{04}
\end{align}

A: J.G.'s answer is perfectly correct. However, rather than getting lost in symbols, a good thing to remember here is the key intuition behind solving this problem:
Exponentials are the eigenfunctions of the derivative operator
The whole problem is simply a souped up application of the fundamental equation $\mathrm{d}_x  \exp(x) = \exp(x)$; now simply use this together with a comparison of the the mnemonics for the curl and cross product:
$$\mathbf{u\times v} = \begin{vmatrix}
  \mathbf{e}_x&\mathbf{e}_y&\mathbf{e}_z\\
  u_1&u_2&u_3\\
  v_1&v_2&v_3\\
\end{vmatrix};\quad \mathbf{\nabla\times v} = \begin{vmatrix}
  \mathbf{e}_x&\mathbf{e}_y&\mathbf{e}_z\\
  \partial_1&\partial_2&\partial_3\\
  v_1&v_2&v_3\\
\end{vmatrix}$$
and your formula is immediately true by inspection, because $-i\,\vec{k}_j$ replaces $\partial_j$, by dint of the exponential's defining property cited above

Note my notation for the basis vectors and Jerry Schirmer's comment:


Why on earth would anyone use $\left({\vec i}, {\vec j}, {\vec k}\right)$
   for the basis vectors and $j$ for $\sqrt{-1}$? You're just begging to confuse people.

