Why is the trace of the outer product of two states equal to the inner product of the two states? Why is it that, given two quantum states $|\psi_1\rangle$, $|\psi_2\rangle$,
$$\mathrm{Tr}(|\psi_1\rangle\langle\psi_2|) = \langle\psi_2|\psi_1\rangle \quad $$
I went through the equation with the qubit space $\mathcal{H}=\{|0\rangle,|1\rangle\}$ and saw that the maths checks out, but I'm lacking any deeper understanding or intuition.
 A: Traces are invariant under cyclical rotation. I.e.
\begin{align}
\mathrm{Tr}(ABCD) = \mathrm{Tr}(DABC) = \mathrm{Tr}(CDAB) = \mathrm{Tr}(BCDA)
\end{align}
This holds for any number of matrices, and the individual maatrices do NOT need to be square. Just their product must be square.
In the physics setting you should interpret $|\psi_1\rangle$ as a column vector (i.e. a tall, narrow matrix), and $\langle\psi_2|$ as a row vector (i.e. a flat, wide matrix). Furthermore $\langle\psi_2|\psi_1\rangle$ is just a number (i.e. a $1\times 1$ matrix), so we can write
\begin{align}
\mathrm{Tr}(|\psi_1\rangle\langle\psi_2|) = \mathrm{Tr}(\langle\psi_2|\psi_1\rangle)=\langle\psi_2|\psi_1\rangle
\end{align}
It is actually a somewhat common trick in physics (for example when computing Feynman diagrams in QFT), to surround a single number with a $\mathrm{Trace}()$ and then cycle the terms around.
A: The calculation itself takes one line. Letting $|n\rangle$ be an arbitrary orthonormal basis, we have
$$\mathrm{Tr}\left(|\psi_1\rangle\langle \psi_2|\right)= \sum_n \langle n|\psi_1\rangle\langle \psi_2|n\rangle = \sum_n\langle \psi_2|n\rangle\langle n|\psi_1\rangle$$
$$ = \langle \psi_2|\left( \sum_n |n\rangle\langle n| \right)|\psi_1\rangle = \langle \psi_2|\psi_1\rangle$$
(Okay, two lines for readability.)
As far as intuition goes, the trace of an operator is equal to the sum of its eigenvalues. The eigenvalue equation for this operator is
$$|\psi_1\rangle\langle \psi_2|\phi\rangle = \lambda |\phi\rangle \implies \langle \psi_2|\psi_1\rangle\langle \psi_2|\phi\rangle = \lambda \langle \psi_2|\phi\rangle$$
$$\implies \bigg(\langle \psi_2|\psi_1\rangle -\lambda\bigg) \langle \psi_2|\phi\rangle  = 0$$
So either $\lambda = \langle \psi_2|\psi_1\rangle \implies |\phi\rangle = |\psi_1\rangle$ or $\langle \psi_2|\phi\rangle = 0$, in which case $\lambda = 0$.  We can therefore conclude that if $|\psi_1\rangle$ and $|\psi_2\rangle$ are orthogonal, then all of the operator's eigenvalues (and therefore its trace) are zero; if they are not orthogonal, then the operator's single non-zero eigenvalue (and again, therefore its trace) is equal to $\langle \psi_2|\psi_1\rangle$.
