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  "body": "\n## Proof\n\nWe show that the permutations of $X$ are exactly the injections from $X$ to $X$, and then count injections.\n\n1. A [[permutation|0BQMWCNS8S]] of $X$ is a [[bijection|0SNQEF7Z53]] $\\sigma : X \\to X$. In particular it is [[injective|0PVG1CRAJN]].\n2. Conversely, let $\\sigma : X \\to X$ be injective, and put $C = \\sigma(X)$. Then $\\sigma$ is a bijection from $X$ onto $C$.\n3. Composing a numbering of $X$ with $\\sigma$ numbers $C$. So $|C| = |X| = n$, as in the [[bijection principle|0GV2573FP4]].\n4. By the [[complement principle|03QC59M99B]], $|X \\setminus C| = n - n = 0$. So $X \\setminus C$ is empty and $C = X$.\n5. So $\\sigma$ is surjective, hence a bijection of $X$, that is, a permutation.\n6. By steps 1 to 5, the permutations of $X$ are exactly the injections from $X$ to $X$.\n7. By the [[number of injections|0X0N833FF8]], there are $(n)_n$ of them, where $(n)_n$ is the [[falling factorial|0N5M924N3P]].\n8. For $n = 0$, $(0)_0 = 1 = 0!$. For $n \\ge 1$, $(n)_n = n (n-1) \\cdots 1$, which is $n!$ by the definition of the [[factorial|0WFCGVMRRC]].\n9. Hence $X$ has exactly $n!$ permutations.\n\n## Idea\n\nOn a finite set, a map from the set to itself that never repeats a value must hit every element.\nSo permutations are just injections from $X$ to itself, and the injection count $n (n-1) \\cdots 1$ is $n!$.\n",
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