Determinant of a product

Key idea. The map $\varphi:B\mapsto\det(AB)$ is multilinear and alternating in the columns of $B$, because $AB$ is linear in the columns of $B$, and the determinant is multilinear and alternating. Moreover, $\varphi(I)=\det(A)$.

Uniqueness. Any alternating multilinear form on $\mathcal{M}_n(\mathbb{R})$ is a scalar multiple of the determinant. So there exists a constant $c$ with $\varphi(B)=c\det(B)$ for all $B$. Setting $B=I$: $\varphi(I)=c\det I=c$, so $c=\det A$.

Conclusion. $\det(AB)=\det A\cdot\det B$ for all $A,B$.

Corollary. If $A$ is invertible, $\det(A^{-1})=1/\det A$. If $A$ is invertible and $B=P^{-1}AP$, then $\det B=\det A$.