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  "body": "\n## Proof\n\n1. Let $N = m + n$ and let $M = (M_{ij})$ be the $N \\times N$ [[block matrix|01AXG9KZQF]] $\\begin{pmatrix} A \u0026 B \\\\ 0 \u0026 D \\end{pmatrix}$.\n2. 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 $i \u003e m$ and $j \\le m$.\n3. By the [[Leibniz formula|016PE4VZ8Q]], $\\det M = \\sum_{\\sigma \\in S_N} \\sgn(\\sigma) \\prod_{i=1}^{N} M_{i, \\sigma(i)}$.\n4. Let $T$ be the set of $\\sigma \\in S_N$ with $\\sigma(i) \u003e m$ for all $i \u003e m$. If $\\sigma \\notin T$, then $\\sigma(i) \\le m$ for some $i \u003e m$, so $M_{i, \\sigma(i)} = 0$ and the term of $\\sigma$ is $0$.\n5. 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.\n6. 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 $k \\le n$.\n7. 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 $S_m \\times S_n$.\n8. 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 $\\sigma = \\alpha' \\beta'$.\n9. 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|0SK90238BH]], $\\sgn(\\alpha') = \\sgn(\\alpha)$.\n10. 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 $1$.\n11. Since $N \\ge 2$, the sign is a [[homomorphism|0JEJJCB0EX]] on $S_N$. So $\\sgn(\\sigma) = \\sgn(\\alpha') \\sgn(\\beta') = \\sgn(\\alpha) \\sgn(\\beta)$.\n12. By step 2, the product in the term of $\\sigma \\in T$ is $\\prod_{i=1}^{m} a_{i, \\alpha(i)} \\cdot \\prod_{k=1}^{n} d_{k, \\beta(k)}$.\n13. By steps 4, 7, 11 and 12, $\\det M = \\sum_{\\alpha \\in S_m} \\sum_{\\beta \\in S_n} \\sgn(\\alpha) \\prod_{i=1}^{m} a_{i, \\alpha(i)} \\cdot \\sgn(\\beta) \\prod_{k=1}^{n} d_{k, \\beta(k)}$.\n14. 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 $\\det D$.\n15. So $\\det M = \\det A \\cdot \\det D$.\n\n## Idea\n\nThe zero block kills every Leibniz term whose permutation sends a lower row to a left column. The remaining permutations keep the two blocks of indices separate, so they split into a permutation of each block. The sum then factors into the Leibniz formulas for $A$ and $D$.\n",
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