Sylvester's determinant identity

Block matrix. Consider
$$M = \begin{pmatrix} I_m & A \\ -B & I_n \end{pmatrix}.$$

First computation. Use the block determinant formula for $\begin{pmatrix} P & Q \\ R & S \end{pmatrix}$ with $P$ invertible: $\det M = \det P\cdot\det(S-RP^{-1}Q)$. Here $P=I_m$, so
$$\det M = \det(I_n - (-B)I_m^{-1}A) = \det(I_n + BA).$$

Second computation. Alternatively, use the formula with $S=I_n$ invertible: $\det M = \det S\cdot\det(P-QS^{-1}R) = \det(I_m - A(-B)) = \det(I_m + AB)$.

Conclusion. $\det(I_m+AB)=\det(I_n+BA)$.