= Solution
For $A=(a_{ij})\in M_n(\mathbb C)$, the <determinant> is
$$
\det A
=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)
\prod_{i=1}^na_{i,\sigma(i)}.
$$
For
$$
A=\begin{pmatrix}M&X\\0&N\end{pmatrix},
$$
any nonzero term in the <Leibniz formula for determinants> must match every row in the lower block to a column in the $N$ block. The remaining upper rows must then match the $M$ columns. The sum consequently factors into the determinant sums for the two diagonal blocks:
$$
\boxed{\det A=(\det M)(\det N)}.
$$
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