= Solution
Use <block matrix> elimination with the invertible square block $A$:
$$
\begin{pmatrix}I&0\\-CA^{-1}&I\end{pmatrix}\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&B\\0&D-CA^{-1}B\end{pmatrix}.
$$
The left multiplier is block triangular with identity diagonal blocks, so its <determinant> is $1$. Taking the <determinant> of the block-triangular product proves
$$
\boxed{\det M=\det A\,\det(D-CA^{-1}B).}
$$
The <matrix> $D-CA^{-1}B$ is the <Schur complement> of $A$. The proof needs $A$ invertible, but does not assume the <Schur complement> or the full <matrix> is invertible.
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