Proof of the determinant of a block triangular matrix by the Leibniz formula

Proof

1. Let $N = m + n$ and let $M = (M_{ij})$ be the $N \times N$ block matrix $\begin{pmatrix} A & B \\ 0 & D \end{pmatrix}$.
2. Then $M_{ij} = a_{ij}$ for $i, j \le m$, and $M_{m+k, m+l} = d_{kl}$ for $k, l \le n$. Also $M_{ij} = 0$ when $i > m$ and $j \le m$.
3. By the Leibniz formula, $\det M = \sum_{\sigma \in S_N} \sgn(\sigma) \prod_{i=1}^{N} M_{i, \sigma(i)}$.
4. Let $T$ be the set of $\sigma \in S_N$ with $\sigma(i) > m$ for all $i > m$. If $\sigma \notin T$, then $\sigma(i) \le m$ for some $i > m$, so $M_{i, \sigma(i)} = 0$ and the term of $\sigma$ is $0$.
5. Let $\sigma \in T$. Then $\sigma$ maps the finite set $\set{m+1, \dots, N}$ injectively into itself, so onto itself. Hence $\sigma$ also maps $\set{1, \dots, m}$ onto itself.
6. So each $\sigma \in T$ gives $\alpha \in S_m$ and $\beta \in S_n$ by $\alpha(i) = \sigma(i)$ for $i \le m$ and $\beta(k) = \sigma(m + k) - m$ for $k \le n$.
7. Conversely, each pair $(\alpha, \beta) \in S_m \times S_n$ comes from exactly one $\sigma \in T$, defined by these two formulas. So $\sigma \mapsto (\alpha, \beta)$ is a bijection from $T$ onto $S_m \times S_n$.
8. Let $\alpha'$ act as $\alpha$ on $\set{1, \dots, m}$ and fix the other points. Let $\beta'$ fix $\set{1, \dots, m}$ and send $m + k$ to $m + \beta(k)$. Then $\sigma = \alpha' \beta'$.
9. A product of transpositions in $S_m$ that equals $\alpha$ is also a product of transpositions in $S_N$ that equals $\alpha'$. By the definition of the sign, $\sgn(\alpha') = \sgn(\alpha)$.
10. Shifting each transposition by $m$ in the same way gives $\sgn(\beta') = \sgn(\beta)$. If $m = 1$ or $n = 1$, the corresponding permutation is the identity, with sign $1$.
11. Since $N \ge 2$, the sign is a homomorphism on $S_N$. So $\sgn(\sigma) = \sgn(\alpha') \sgn(\beta') = \sgn(\alpha) \sgn(\beta)$.
12. By step 2, the product in the term of $\sigma \in T$ is $\prod_{i=1}^{m} a_{i, \alpha(i)} \cdot \prod_{k=1}^{n} d_{k, \beta(k)}$.
13. By steps 4, 7, 11 and 12, $\det M = \sum_{\alpha \in S_m} \sum_{\beta \in S_n} \sgn(\alpha) \prod_{i=1}^{m} a_{i, \alpha(i)} \cdot \sgn(\beta) \prod_{k=1}^{n} d_{k, \beta(k)}$.
14. The double sum factors as a product of a sum over $\alpha$ and a sum over $\beta$. By the Leibniz formula, these sums are $\det A$ and $\det D$.
15. So $\det M = \det A \cdot \det D$.