Determinant of a block triangular matrix

Reduction to triangular case. Put $A$ in upper triangular form by left multiplication by an invertible matrix $P$ (Gaussian elimination). Applying $P$ block-diagonally as $\begin{pmatrix} P & 0 \\ 0 & I_q \end{pmatrix}$ transforms $M$ into $\begin{pmatrix} PA & PB \\ 0 & C \end{pmatrix}$ without changing the block structure, and $\det(PA)=\det P\det A$. Similarly reduce $C$ to upper triangular form. So we may assume $A$ and $C$ are upper triangular.

Triangular case. In this case $M$ is upper triangular with diagonal entries those of $A$ and $C$, so
$$\det M = \prod_i a_{ii}\cdot\prod_j c_{jj} = \det A\cdot\det C.$$

Conclusion. The formula holds in general.