Invariance of eigenvalues of a product of square matrices under cyclic permutation I recently came across this proposition that the eigenvalues of a product of square matrices are invariant under cyclic permutation of the product order. Is there perhaps some group theoretic way of proving this proposition? 
I've tried a few cases and it seems to be true, but a direct proof has proven elusive. Or is a proof not so simple?
 A: With only a little extra work we can generalize this result to rectangluar matrices, as long as their sizes are such that cyclic permutation is well defined.
Specifically, suppose $A$ is a wide $r$-by-$n$ matrix, and $B$ is a tall $n$-by-$r$ matrix. The nonzero eigenvalues of the $r$-by-$r$ matrix $AB$ are the same as the eigenvalues of a larger $n$-by-$n$ matrix formed by padding with blocks of zeros. 
$$\lambda_\text{nonzero}(AB) = \lambda_\text{nonzero}\left(\underbrace{\begin{bmatrix}AB & 0 \\ 0 & 0\end{bmatrix}}_{n\times n\text{ square matrix}}\right)$$
This larger matrix can then be factored as follows:
$$\begin{bmatrix}AB & 0 \\ 0 & 0\end{bmatrix} = \underbrace{ \begin{bmatrix}A \\ 0\end{bmatrix}}_{n\times n\text{ square matrix}}\underbrace{\begin{bmatrix}B & 0\end{bmatrix}}_{n\times n\text{ square matrix}}.$$
But by the result for square matrices (explained by EuYu), the nonzero eigenvalues are preserved if we commute these square factors. Doing so leads to,
$$\begin{bmatrix}B & 0\end{bmatrix}\begin{bmatrix}A \\ 0\end{bmatrix} = BA + 0 = BA.$$
Hence the nonzero eigenvalues of $AB$ equal the nonzero eigenvalues of $BA$.
This can then be extended to cyclic permutations of rectangular matrices that are permitted by their sizes, using the same argument as EuYu explained for square matrices.
A: This is a consequence of the fact that the spectrum of $AB$ and $BA$ coincide. Suppose that $\lambda$ is a non-zero eigenvalue of $AB$ with eigenvector $\mathbf{v}$. Then
$$AB\mathbf{v} = \lambda\mathbf{v}\implies BA(B\mathbf{v}) = \lambda B\mathbf{v}$$
so that $B\mathbf{v}$ is an eigenvector of $\lambda$ for $BA$. Clearly $AB$ and $BA$ are either both invertible or non-invertible so $0$ as an eigenvalue also coincides. 
Therefore $\sigma(AB) = \sigma(BA)$ where $\sigma$ denotes the spectrum.
This fact then extends to any cyclic permutation (much like how the cyclic permutation property of the trace also follows from $\mathrm{tr}(AB) = \mathrm{tr}(BA)$). For example, by continuously applying the exchange property to this cluster of $4$ matrices, we have
$$\sigma([ABC]D) = \sigma(D[ABC]) = \sigma([DAB]C) = \sigma(C[DAB]) = \cdots  $$
