= Solution
In the <determinant> expansion of the <block upper triangular matrix>, a nonzero product must assign each of the bottom $n$ rows to one of the last $n$ columns, because the lower-left block is zero. Those rows therefore occupy all the last $n$ columns, leaving the first $n$ columns for the top rows. The contributing permutations are precisely independent permutations within the two blocks. Their signs multiply, so the sum factors:
$$
\boxed{\det\begin{pmatrix}A&C\\0&B\end{pmatrix}=\det A\,\det B.}
$$
The off-diagonal block contributes to no surviving term.
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