Suppose that one has two matrices $A$, $B$. Then
Prove that $$|AB - \lambda I| = |BA - \lambda I|,$$ where $|\cdot|$ denotes the determinant, $I$ - identity matrix and $\lambda \in \mathbb{C}$.
Note that $A$ and $B$ are not necessary invertible. For invertible matrices I easily found $$|AB - \lambda I| = |B(AB - \lambda I)B^{-1}| = |BA - \lambda I|.$$
On
There is an old-fashioned proof depending only on the properties of determinants and minors.
(i) In $\det(xI-X)$ the coefficient of $x^{n-k}$ is the sum of the principal $k\times k$ minors of $X$.
(ii) Let $X^{(k)}$ denote the matrix of $k\times k$ minors of $X$ for any square $X$. Then by a theorem sometimes called the Binet-Gauss Theorem we have that $(XY)^{(k)}=X^{(k)}Y^{(k)}$.
(iii) For any square matrices $X,Y$ we have $\text{tr}(XY)=\text{tr}(YX)$.
So the coefficient of $x^{n-k}$ in $\det(xI-AB)$ is $$\text{tr}((AB)^{(k)})=\text{tr}(A^{(k)}B^{(k)})$$ whereas the coefficient of $x^{n-k}$ in $\det(xI-BA)$ is $$\text{tr}((BA)^{(k)})=\text{tr}(B^{(k)}A^{(k)})$$ and these are equal.
I'm aware of three proofs. The first is the one you get by combining your argument with Marko's comment. The second is to show that $AB$ and $BA$ have the same non-zero eigenvalues with the same multiplicities. The third proof runs as follows. Set \[ M=\begin{pmatrix}I&A\\ B&I\end{pmatrix},\quad N=\begin{pmatrix}I&0\\-B&I\end{pmatrix} \] and then note that \[ MN=\begin{pmatrix}I-AB&A\\0&I\end{pmatrix},\quad NM=\begin{pmatrix}I&A\\0&I-BA\end{pmatrix}. \] Since $\det(MN)=\det(NM)$, the result follows.
One advantage of the third proof is that it works for matrices over a commutative ring.