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  "body": "\n## Proof\n\nWe apply the [[inclusion–exclusion principle|0FF7726C7P]] to the permutations that fix a given point.\n\n1. Let $X$ be a set with $n$ elements. Choose a bijection $\\beta : X \\to [n]$, where $[n] = \\set{1, \\dots, n}$.\n2. The map $\\sigma \\mapsto \\beta \\sigma \\beta^{-1}$ is a bijection from the [[permutations|0BQMWCNS8S]] of $X$ to those of $[n]$. Its inverse is $\\tau \\mapsto \\beta^{-1} \\tau \\beta$.\n3. We have $\\beta \\sigma \\beta^{-1}(\\beta(x)) = \\beta(x)$ exactly when $\\sigma(x) = x$. So this bijection sends [[derangements|0TH61JR04G]] to derangements, in both directions.\n4. By the [[bijection principle|0GV2573FP4]], it suffices to count the derangements of $[n]$.\n5. If $n = 0$, the only permutation of $[0] = \\varnothing$ is the empty map. It has no fixed point, so $D_0 = 1$. The formula gives $0! \\cdot \\frac{1}{0!} = 1$. From now on let $n \\ge 1$.\n6. Let $\\Sigma$ be the set of permutations of $[n]$. By the [[number of permutations|0CEBCS0E0E]], $|\\Sigma| = n!$.\n7. For $i \\in [n]$, let $A_i = \\set{\\sigma \\in \\Sigma : \\sigma(i) = i}$. A permutation is a derangement exactly when it lies in no $A_i$.\n8. So the derangements form the complement of $\\bigcup_{i=1}^{n} A_i$ in $\\Sigma$. By the [[complement principle|03QC59M99B]], $D_n = n! - \\left| \\bigcup_{i=1}^{n} A_i \\right|$.\n9. Let $I \\subseteq [n]$ have $k$ elements, and let $A_I = \\bigcap_{i \\in I} A_i$, the permutations that fix every point of $I$. We claim $|A_I| = (n - k)!$.\n10. Let $\\sigma \\in A_I$ and $x \\notin I$. If $\\sigma(x) = i \\in I$, then $\\sigma(x) = \\sigma(i)$ with $x \\ne i$, against injectivity. So $\\sigma$ maps $[n] \\setminus I$ into itself.\n11. This restriction is injective. It is also surjective: if $y \\notin I$ and $y = \\sigma(x)$, then $x \\notin I$, since otherwise $y = \\sigma(x) = x \\in I$.\n12. So restriction is a map from $A_I$ to the permutations of $[n] \\setminus I$. Its inverse extends a permutation of $[n] \\setminus I$ by the identity on $I$.\n13. Hence $A_I$ is in bijection with the permutations of an $(n - k)$-element set. By the number of permutations, $|A_I| = (n - k)!$.\n14. By the [[number of $k$-element subsets|0BFMR2FQ8B]], there are $\\binom{n}{k}$ sets $I$ with $k$ elements. By the [[factorial formula|033KTKKRV6]], $\\binom{n}{k} (n - k)! = \\frac{n!}{k!}$.\n15. The inclusion–exclusion principle, with terms grouped by $k = |I|$, gives\n$$\\left| \\bigcup_{i=1}^{n} A_i \\right| = \\sum_{k=1}^{n} (-1)^{k+1} \\binom{n}{k} (n - k)! = \\sum_{k=1}^{n} (-1)^{k+1} \\frac{n!}{k!}.$$\n16. By step 8, and since $n! = (-1)^0 \\frac{n!}{0!}$ by the definition of the [[factorial|0WFCGVMRRC]],\n$$D_n = n! - \\sum_{k=1}^{n} (-1)^{k+1} \\frac{n!}{k!} = \\sum_{k=0}^{n} (-1)^k \\frac{n!}{k!} = n! \\sum_{k=0}^{n} \\frac{(-1)^k}{k!}.$$\n\n## Idea\n\nA derangement is a permutation that avoids every set $A_i$ of permutations fixing $i$. The permutations that fix $k$ given points are just the permutations of the other $n - k$ points, so inclusion–exclusion has easy terms. They simplify to $(-1)^k n!/k!$.\n",
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