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  "body": "\n## Proof\n\n1. Let $A = (a_{ij})$. By the definition of the [[transpose|0W809K6AW0]], $(A^T)_{ij} = a_{ji}$.\n2. By the [[Leibniz formula|016PE4VZ8Q]] applied to $A^T$, $\\det A^T = \\sum_{\\sigma \\in S_n} \\sgn(\\sigma) \\prod_{i=1}^{n} a_{\\sigma(i), i}$.\n3. Fix $\\sigma$. As $i$ runs over $1, \\dots, n$, the index $k = \\sigma(i)$ also runs over $1, \\dots, n$, and $i = \\sigma^{-1}(k)$.\n4. Since $K$ is commutative, we may reorder the product. So $\\prod_{i=1}^{n} a_{\\sigma(i), i} = \\prod_{k=1}^{n} a_{k, \\sigma^{-1}(k)}$.\n5. If $n \\ge 2$, the sign is a [[homomorphism|0JEJJCB0EX]]. So $\\sgn(\\sigma)\\sgn(\\sigma^{-1}) = \\sgn(\\sigma \\sigma^{-1}) = 1$.\n6. Both signs lie in $\\set{1, -1}$, so step 5 gives $\\sgn(\\sigma^{-1}) = \\sgn(\\sigma)$. If $n = 1$, then $\\sigma$ is the identity and this is clear.\n7. By steps 2, 4 and 6, $\\det A^T = \\sum_{\\sigma \\in S_n} \\sgn(\\sigma^{-1}) \\prod_{k=1}^{n} a_{k, \\sigma^{-1}(k)}$.\n8. The map $\\sigma \\mapsto \\sigma^{-1}$ is a bijection of the [[symmetric group|0BQMWCNS8S]] $S_n$ onto itself. So we may write $\\rho = \\sigma^{-1}$ and sum over $\\rho \\in S_n$.\n9. This gives $\\det A^T = \\sum_{\\rho \\in S_n} \\sgn(\\rho) \\prod_{k=1}^{n} a_{k, \\rho(k)} = \\det A$, by the Leibniz formula for $A$.\n\n## Idea\n\nThe term of a permutation $\\sigma$ for $A^T$ uses the same entries as the term of $\\sigma^{-1}$ for $A$. A permutation and its inverse have the same sign, so the two Leibniz sums have the same terms in a different order.\n",
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