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  "body": "\n## Proof\n\nWe expand each row in the standard basis and use the properties from the [[definition of the determinant|0C3CFHDN6R]].\n\n1. Let $e_1, \\dots, e_n$ be the rows of the [[identity matrix|06DW9BAWB1]] $I_n$. So $e_j$ has entry $1$ in place $j$ and $0$ elsewhere.\n2. Row $i$ of $A$ is $\\sum_{j=1}^{n} a_{ij} e_j$.\n3. For a map $f \\colon \\set{1, \\dots, n} \\to \\set{1, \\dots, n}$, let $E_f$ be the matrix whose row $i$ is $e_{f(i)}$.\n4. Apply [[multilinearity|0PPG793642]] in row $1$ to the sum of step 2. Then apply it in row $2$ to each resulting term, and so on up to row $n$.\n5. This gives $\\det A = \\sum_{f} \\left(\\prod_{i=1}^{n} a_{i, f(i)}\\right) \\det E_f$. The sum runs over all $n^n$ maps $f$ of step 3.\n6. If $f$ is not injective, then $f(i) = f(k)$ for some $i \\ne k$. Then rows $i$ and $k$ of $E_f$ are equal, so $\\det E_f = 0$ by the rule for [[two equal rows|0K4VQB7HWR]].\n7. An injective map from the finite set $\\set{1, \\dots, n}$ to itself is a bijection. So the remaining maps $f$ are exactly the [[permutations|05RWFKVTJ7]] $\\sigma \\in S_n$.\n8. For $\\sigma \\in S_n$, row $i$ of $E_\\sigma$ is row $\\sigma(i)$ of $I_n$. By the rule for [[permuting rows|09YB3WQG38]], $\\det E_\\sigma = \\sgn(\\sigma) \\det I_n$.\n9. By the [[determinant of the identity|0Z4BWARC38]], $\\det I_n = 1$. So $\\det E_\\sigma = \\sgn(\\sigma)$.\n10. Putting steps 5 to 9 together, $\\det A = \\sum_{\\sigma \\in S_n} \\sgn(\\sigma) \\prod_{i=1}^{n} a_{i, \\sigma(i)}$.\n\n## Idea\n\nEach row is a combination of the standard rows, so multilinearity breaks $\\det A$ into $n^n$ simple pieces. A piece with a repeated standard row is zero. The others are row permutations of the identity, whose determinant is the sign of the permutation.\n",
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