Multiplicativity of the determinant

First recall the effect of the three elementary row operations on the determinant (established directly from the Leibniz sum $\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}$, since each term of that sum is linear in each row separately): swapping two rows multiplies $\det$ by $-1$, scaling one row by $k$ multiplies $\det$ by $k$, and adding a multiple of one row to another leaves $\det$ unchanged.

Each elementary row operation on $A$ is the same as left-multiplying $A$ by a corresponding elementary matrix $E$ (obtained by applying that same operation to the identity matrix $I$). Comparing with the previous paragraph, $\det(E)$ equals exactly the scaling factor of that operation ($-1$, $k$, or $1$ respectively), so for every elementary matrix $E$, $\det(EA)=\det(E)\det(A)$.

Case $A$ invertible: Gaussian elimination reduces any invertible matrix to the identity using a finite sequence of elementary row operations, i.e. $E_m\cdots E_1A=I$ for elementary matrices $E_1,\dots,E_m$, so $A=E_1^{-1}\cdots E_m^{-1}$, itself a product of elementary matrices (the inverse of an elementary matrix is again elementary, of the same type). Applying the previous paragraph's identity repeatedly, $\det(A)=\det(E_1^{-1})\cdots\det(E_m^{-1})$, and applying it again to $AB=E_1^{-1}\cdots E_m^{-1}B$ gives $\det(AB)=\det(E_1^{-1})\cdots\det(E_m^{-1})\det(B)=\det(A)\det(B)$.

Case $A$ singular: then $\det(A)=0$ (a singular matrix cannot be reduced all the way to the identity, and the row-operation rules above show every reachable row-echelon form still has a zero row, forcing $\det=0$ by cofactor expansion along that row). Singularity also means $\operatorname{rank}(A)<n$, and since $\operatorname{rank}(AB)\leq\operatorname{rank}(A)<n$ for any $B$, the product $AB$ is singular too, so $\det(AB)=0$ as well. Hence $\det(AB)=0=0\cdot\det(B)=\det(A)\det(B)$, and the identity holds in this case too, completing the proof for every $A$.