Let $A \in M_{mxn}(F)$ and $B \in M_{nxm}(F)$, show that $$x^ndet|xI_m-AB| = x^mdet|x I-BA|.$$ Deduce that if n=m, then AB and BA have the same characteristic polynomial.
The second part should be easy to prove: if n=m, then $$x^ndet|xI-AB| = x^ndet|x I-BA| \implies det|xI-AB| = det|xI-BA|$$ which means the characteristic polynomial is the same since $det|xI-AB|$ or $det|xI-BA|$ gives the exact characteristic polynomial of AB and BA respectively.
For the first part, I am not sure how to proceed with the left side being determinant of mxm matrix while right side being determinant of nxn matrix.
Consider the matrices $\mathcal C$, $\mathcal D$ below, compute $\mathcal C\mathcal D$ and $\mathcal D\mathcal C$ $$ \begin{aligned} \mathcal C&=\begin{bmatrix} I_m & -A\\ & I_n\end{bmatrix}\ ,\\ \mathcal D&=\begin{bmatrix} xI_m & xA\\ B&xI_n\end{bmatrix}\ ,\\[3mm] % \mathcal C\mathcal D &= \begin{bmatrix} I_m & -A\\ & I_n\end{bmatrix} \begin{bmatrix} xI_m & xA\\ B&xI_n\end{bmatrix} = \begin{bmatrix} xI_m-AB & \\ *& xI_n\end{bmatrix}\ ,\\ % \mathcal D\mathcal C &= \begin{bmatrix} xI_m & xA\\ B&xI_n\end{bmatrix} \begin{bmatrix} I_m & -A\\ & I_n\end{bmatrix} = \begin{bmatrix} xI_m & \\ * & xI_n-BA\end{bmatrix}\ , \end{aligned} $$ (empty entries are zero block matrices, star entries do not matter any more,) and use finally the multiplicativity of the determinant $$ \det(\mathcal C\mathcal D) =\det(\mathcal C)\det(\mathcal D) =\det(\mathcal D)\det(\mathcal C) =\det(\mathcal D\mathcal C) \ , $$ and the relations $$ \det\begin{bmatrix}S&\\*&T\end{bmatrix}=\det(S)\det(T)\ ,\qquad \det(xI_m)=x^m\det(I_m)=x^m\ . $$