How to show if $ \lambda$ is an eigenvalue of $AB^{-1}$, then $ \lambda$ is an eigenvalue of $ B^{-1}A$?

A shorter way of seeing this would be to observe that if $$ (AB^{-1})x=\lambda x $$ for some non-zero vector $x$, then by multiplying that equation by $B^{-1}$ (from the left) we get that $$ (B^{-1}A)(B^{-1}x)=\lambda (B^{-1}x). $$ In other words $(B^{-1}A)y=\lambda y$ for the non-zero vector $y=B^{-1}x$. This process is clearly reversible.

Even if $A$ is $n\times m$ and $B$ is $m\times n$ with $m\le n$, we have $$ \det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)\tag{1} $$ Proof:

Drawing from an answer of julien's, $$ \begin{bmatrix}I_n&-A\\0&\lambda I_m\end{bmatrix} \begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix} =\begin{bmatrix}\lambda I_n-AB&0\\\lambda B&\lambda I_m\end{bmatrix}\tag{2} $$ $$ \begin{bmatrix}I_n&0\\-B&\lambda I_m\end{bmatrix} \begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix} =\begin{bmatrix}\lambda I_n&A\\0&\lambda I_m-BA\end{bmatrix}\tag{3} $$ Since the determinants on the left sides of $(2)$ and $(3)$ are equal, the determinants on the right side prove $$ \lambda^m\det(\lambda I_n-AB)=\lambda^n\det(\lambda I_m-BA)\tag{4} $$ In the case of square matrices, since the characteristic polynomials are the same, the eigenvalues are the same.