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  "body": "\n## Proof\n\n1. Let $n$ be a positive integer and let $X = \\{1, 2, \\dots, n\\}$, a set with $n$ elements.\n2. By the definition of the [[symmetric group|0RJMNEN3BJ]], $S_n = \\operatorname{Sym}(X)$ is the set of all bijections from $X$ to $X$.\n3. These bijections are exactly the [[permutations|0BQMWCNS8S]] of $X$.\n4. By the [[number of permutations|0CEBCS0E0E]], $X$ has exactly $n!$ permutations.\n5. The [[order|0SEX9F9J5Z]] of $S_n$ is the number of its elements. By steps 2 to 4, it is $n!$.\n\n## Idea\n\nThe group $S_n$ is just the set of permutations of $n$ symbols with composition. So its order is the number of such permutations, which is $n!$ by a counting argument. The image of $1$ has $n$ choices, the image of $2$ then has $n - 1$ choices, and so on.\n",
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