= Solution
Use the <Leibniz formula for determinants>:
$$
\det M=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{j=1}^n m_{j,\sigma(j)}.
$$
For a nonzero product in an <upper triangular matrix>, every selected column must satisfy $\sigma(j)\geq j$. But a <permutation> has $\sum_j\sigma(j)=\sum_jj$, so every one of these inequalities is an equality. Only the identity <permutation> can contribute. Its sign is positive and its product is $\prod_jm_{jj}$, which proves the result even if some diagonal entries vanish.
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