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  "body": "\n## Statement\n\nLet $q$ be a prime power, let $\\mathbb{F}_q$ be the finite field with $q$ elements, and let $n \\geq 1$.\nThe group $\\mathrm{GL}_n(\\mathbb{F}_q)$ of the invertible $n \\times n$ matrices over $\\mathbb{F}_q$ has [[order|0B85S2V2DR]] $\\prod_{k=0}^{n-1} (q^n - q^k) = (q^n - 1)(q^n - q) \\cdots (q^n - q^{n-1})$.\n\n## Remarks\n\nA matrix is invertible exactly when its columns are linearly independent.\nThere are $q^n - 1$ choices for the first column. Column $k+1$ has $q^n - q^k$ choices, because it must avoid the $q^k$ vectors in the span of the first $k$ columns.\nFor example, $|\\mathrm{GL}_2(\\mathbb{F}_2)| = 3 \\cdot 2 = 6$, and $\\mathrm{GL}_2(\\mathbb{F}_2) \\cong S_3$.\nAlso, $|\\mathrm{GL}_3(\\mathbb{F}_2)| = 7 \\cdot 6 \\cdot 4 = 168$.\n\n## History\n\nÉvariste Galois computed the order of the general linear group over a prime field in 1832, in his last letter to Chevalier.\n\n## Explanations\n\n### L1 human\n\nTo build an invertible matrix, choose its columns one after the other.\nEach new column must not lie in the span of the columns that you already chose.\nCounting the choices at each step gives the product formula.\n",
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