Prove that $\det(xI_m-AB)=x^{m-n}\det(xI_n-BA)$ I want to prove that
$$\det \left( x I_m - A B \right) = x^{m-n} \det \left( x I_n - B A \right)$$
where $A \in \mathbb{F}^{m \times n}$ and  $B \in \mathbb{F}^{n\times m}$.

It is easy to show that $0$ has algebraic multiplicity of at least $m-n$ for $AB$, but how can I show that the other eigenvalues of $AB$ are actually eigenvalues of $BA$ ? I know that the sum of eigenvalues of $AB$ and $BA$ are the same, because $\operatorname{trace}(AB)=\operatorname{trace}(BA)$, but ...
I appreciate any help, thanks ! ${{{}}}$

This proof is nothing new. It is a reformulation of Sylvester's determinant theorem as suggested in Fabian's comment.
Assume $x \ne 0$, we have
$$
\begin{bmatrix}I_n & B\\ A & xI_m\end{bmatrix}
\begin{bmatrix}I_n & -B\\ 0 & I_m\end{bmatrix} =
\begin{bmatrix}I_n & 0\\ A & xI_m - AB\end{bmatrix}$$
This implies
$$\det(xI_m - AB) =
\begin{vmatrix}I_n & 0\\ A & xI_m - AB\end{vmatrix}
=\begin{vmatrix}I_n & B\\ A & xI_m\end{vmatrix}\tag{*1}
$$
Notice 
$$\begin{vmatrix}I_n & B\\ A & xI_m\end{vmatrix}
= (-)^{mn} \begin{vmatrix}A & xI_m\\I_n & B\end{vmatrix}
= \begin{vmatrix}xI_m & A\\B & I_n\end{vmatrix}
= x^{m+n}\begin{vmatrix}I_m & x^{-1}A\\x^{-1}B & x^{-1}I_n\end{vmatrix}
$$
Apply $(*1)$ with $x^{-1}B, x^{-1}A, x^{-1}I_n$ taking the roles of $A, B, xI_m$, we get
$$\begin{align}\det(xI_m-AB) &= x^{m+n}\det(x^{-1}I_n - x^{-2}BA) = x^{(m+n)-2n}\det(x I_n - BA)\\
&= x^{m-n}\det(xI_n - BA)\end{align}$$
This leave us with the special case $x = 0$, 


*

*if $m = n$, the relation at hand reduces to $\det(-AB) = \det(-BA)$ which is trivially true. 

*if $m > n$, LHS reduces to $\det(-AB)$ where $AB$ is a $m \times m$ matrix. Since the rank of $A$ and $B$ is at most $n < m$, this determinant is zero. RHS is zero because of the $x^{m-n}$ factor.


Combine all these, we have
$$\det(x I_m - AB ) = x^{m-n} \det(x I_n - BA)$$
unless $x = 0$ and $m < n$ where the RHS is an undefined expression.
