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  "body": "\n## Proof\n\nWe expand the rows of $AB$ by multilinearity and sort the terms by the set of rows of $B$ that they use.\n\n1. Let $A = (a_{ik})$ be $m \\times n$ and $B$ be $n \\times m$ over $K$, with $1 \\le m \\le n$. Here $m \\ge 1$, since the [[determinant|0C3CFHDN6R]] is defined for sizes at least $1$. Let $\\beta_1, \\dots, \\beta_n$ be the rows of $B$.\n2. By the definition of the [[matrix product|076JGP6R3T]], row $i$ of the $m \\times m$ matrix $AB$ is $\\sum_{k=1}^{n} a_{ik} \\beta_k$.\n3. For a map $f \\colon \\set{1, \\dots, m} \\to \\set{1, \\dots, n}$, let $B_f$ be the $m \\times m$ matrix whose row $i$ is $\\beta_{f(i)}$.\n4. Apply [[multilinearity in the rows|0PPG793642]] to the sums of step 2, in each of the rows $1, \\dots, m$ in turn.\n5. This gives $\\det(AB) = \\sum_{f} \\left(\\prod_{i=1}^{m} a_{i, f(i)}\\right) \\det B_f$, over all maps $f$ of step 3.\n6. If $f$ is not injective, then $B_f$ has two equal rows. So $\\det B_f = 0$ by the rule for [[two equal rows|0K4VQB7HWR]].\n7. Let $f$ be injective, and let $S = f(\\set{1, \\dots, m})$. Then $S$ has $m$ elements. Write $S = \\set{s_1 \u003c s_2 \u003c \\cdots \u003c s_m}$.\n8. There is exactly one [[permutation|05RWFKVTJ7]] $\\sigma \\in S_m$ with $f(i) = s_{\\sigma(i)}$ for all $i$. Conversely, each pair $(S, \\sigma)$ defines an injective $f$ in this way.\n9. So the injective maps $f$ correspond one to one to the pairs $(S, \\sigma)$, where $S$ has $m$ elements and $\\sigma \\in S_m$.\n10. Row $r$ of $B^S$ is $\\beta_{s_r}$. So row $i$ of $B_f$ is row $\\sigma(i)$ of $B^S$.\n11. By the rule for [[permuting rows|09YB3WQG38]], $\\det B_f = \\sgn(\\sigma) \\det B^S$.\n12. Column $r$ of $A_S$ is column $s_r$ of $A$. So $a_{i, f(i)} = a_{i, s_{\\sigma(i)}} = (A_S)_{i, \\sigma(i)}$.\n13. By steps 5 to 12, $\\det(AB) = \\sum_{S} \\left(\\sum_{\\sigma \\in S_m} \\sgn(\\sigma) \\prod_{i=1}^{m} (A_S)_{i, \\sigma(i)}\\right) \\det B^S$.\n14. By the [[Leibniz formula|016PE4VZ8Q]], the inner sum is $\\det A_S$. So $\\det(AB) = \\sum_{S} \\det(A_S) \\det(B^S)$, over the subsets $S$ with $m$ elements.\n\n## Idea\n\nEach row of $AB$ is a combination of the rows of $B$. Expanding by multilinearity leaves only choices of $m$ distinct rows of $B$. For a fixed set $S$ of rows, the orderings of $S$ give $\\det B^S$ times the Leibniz formula for $\\det A_S$.\n",
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