$A$ is an $m\times n$ matrix and $B$ is an $n\times m$ matrix. $$ \det(I_m + AB) = \det(I_n + BA) $$ Solution: I found this guys: $$ \det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix} \det\begin{pmatrix}I&B\\\\0&I\end{pmatrix} =\det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix}\begin{pmatrix}I&B\\\\0&I\end{pmatrix} =\det\begin{pmatrix}I&0\\\\A&AB+I\end{pmatrix} =\det(I+AB) $$
and
$$ \det\begin{pmatrix}I&B\\\\0&I\end{pmatrix} \det\begin{pmatrix}I&-B\\\\A&I\end{pmatrix} =\det\begin{pmatrix}I&B\\\\0&I\end{pmatrix} \begin{pmatrix}I&-B\\\\A&I\end{pmatrix} =\det\begin{pmatrix}I+BA&0\\\\A&I\end{pmatrix} =\det(I+BA) $$
Source: Sylvester's determinant identity
If $B$ is (square and) invertible, then \begin{align*} \det (1 + AB) = \det \left[B^{-1}(1 + BA)B\right] = (\det B^{-1}) \det(1 + BA)\det B) = \det (1 + BA). \end{align*} But $f(A, B) = \det (1 + AB) - \det (1 + BA)$ is a polynomial in the entries $A_{ij}$ and $B_{ij}$, and the space of invertible $B$ is dense (as if $B$ is not invertible, then $B + \epsilon $ is invertible for small $\epsilon > 0$). Hence $f(A, B)$ vanishes everywhere, as required. For the arbitrary case, pad $A$ and $B$ with $0$s and reduce to the square case.
(That's assuming that you're working over a ground field like $\mathbb{R}$ or $\mathbb{C}$. To make the argument work in completely generality, you'll either need to be more precise about the nature of the "dense set," embed $A$ and $B$ in a larger $GL_n(k)$ as in the wiki link above, explicitly work with something like a $QR$-decomposition of $A$ and $B$, or etc.)