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  "body": "\n## Proof\n\nWe expand the rows of $AB$ by multilinearity, as in the proof of the Leibniz formula.\n\n1. Let $A = (a_{ij})$ and let $\\beta_1, \\dots, \\beta_n$ be the rows of $B = (b_{kj})$.\n2. By the definition of the [[matrix product|076JGP6R3T]], $(AB)_{ij} = \\sum_{k=1}^{n} a_{ik} b_{kj}$. So row $i$ of $AB$ is $\\sum_{k=1}^{n} a_{ik} \\beta_k$.\n3. For a map $f \\colon \\set{1, \\dots, n} \\to \\set{1, \\dots, n}$, let $B_f$ be the matrix whose row $i$ is $\\beta_{f(i)}$.\n4. Apply [[multilinearity|0PPG793642]] in each of the rows $1, \\dots, n$ in turn to the sums of step 2.\n5. This gives $\\det(AB) = \\sum_{f} \\left(\\prod_{i=1}^{n} a_{i, f(i)}\\right) \\det B_f$, 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. The injective maps $f$ are bijections, so they are the [[permutations|05RWFKVTJ7]] $\\sigma \\in S_n$.\n8. For $\\sigma \\in S_n$, row $i$ of $B_\\sigma$ is row $\\sigma(i)$ of $B$. By the rule for [[permuting rows|09YB3WQG38]], $\\det B_\\sigma = \\sgn(\\sigma) \\det B$.\n9. By steps 5 to 8, $\\det(AB) = \\left(\\sum_{\\sigma \\in S_n} \\sgn(\\sigma) \\prod_{i=1}^{n} a_{i, \\sigma(i)}\\right) \\det B$.\n10. By the [[Leibniz formula|016PE4VZ8Q]], the sum in step 9 is $\\det A$. So $\\det(AB) = \\det A \\cdot \\det B$.\n\n## Idea\n\nEach row of $AB$ is a combination of the rows of $B$ with coefficients from $A$. Expanding by multilinearity leaves only the row permutations of $B$, each with determinant $\\pm \\det B$. The coefficients that remain add up to the Leibniz formula for $\\det A$.\n",
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