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  "body": "\n## Proof\n\nWe sort the derangements by the image of the last point and count each class by a bijection.\n\n1. Conjugating by a bijection between two sets with $m$ elements matches their [[derangements|0TH61JR04G]]. So by the [[bijection principle|0GV2573FP4]], $D_m$ depends only on $m$.\n2. Let $n$ be a natural number, let $X = \\set{1, \\dots, n + 2}$, let $z = n + 2$, and let $Y = X \\setminus \\set{z}$. So $|Y| = n + 1$.\n3. Let $\\sigma$ be a derangement of $X$. Then $\\sigma(z) \\ne z$, so $\\sigma(z) \\in Y$.\n4. Fix $i \\in Y$. Let $E_i$ be the set of derangements $\\sigma$ of $X$ with $\\sigma(z) = i$. Split $E_i$ into $E_i'$, where $\\sigma(i) = z$, and $E_i''$, where $\\sigma(i) \\ne z$.\n5. Count $E_i'$. Let $Z = X \\setminus \\set{i, z}$, a set with $n$ elements. For $\\sigma \\in E_i'$, $\\sigma$ maps $\\set{i, z}$ onto itself, so its restriction to $Z$ is a [[permutation|0BQMWCNS8S]] of $Z$.\n6. This restriction has no fixed point, since $\\sigma$ has none. Conversely, a derangement of $Z$ extended by $i \\mapsto z$ and $z \\mapsto i$ lies in $E_i'$.\n7. These two maps are inverse to each other. So $|E_i'| = D_n$.\n8. Count $E_i''$. For $\\sigma \\in E_i''$, let $j = \\sigma^{-1}(z)$. Then $j \\ne z$ because $\\sigma(z) = i \\ne z$, and $j \\ne i$ because $\\sigma(i) \\ne z$.\n9. Define $\\tau : Y \\to Y$ by $\\tau(j) = i$ and $\\tau(x) = \\sigma(x)$ for $x \\in Y \\setminus \\set{j}$. These values lie in $Y$, since only $j$ is sent to $z$ by $\\sigma$.\n10. The map $\\tau$ is a bijection of $Y$. Indeed, $\\sigma$ maps $Y \\setminus \\set{j}$ onto $X \\setminus \\set{z, i}$, since $\\sigma(j) = z$ and $\\sigma(z) = i$. Adding $\\tau(j) = i$ gives all of $Y$.\n11. The map $\\tau$ has no fixed point. For $x \\ne j$, $\\tau(x) = \\sigma(x) \\ne x$. Also $\\tau(j) = i \\ne j$ by step 8.\n12. Conversely, let $\\tau$ be a derangement of $Y$, and let $j = \\tau^{-1}(i)$. Define $\\sigma$ on $X$ by $\\sigma(z) = i$, $\\sigma(j) = z$, and $\\sigma(x) = \\tau(x)$ for the other $x$.\n13. This $\\sigma$ is a permutation of $X$, because $\\tau$ maps $Y \\setminus \\set{j}$ onto $Y \\setminus \\set{i}$, and $\\sigma$ adds $j \\mapsto z$ and $z \\mapsto i$. It has no fixed point: $\\sigma(z) = i \\ne z$, $\\sigma(j) = z \\ne j$, and $\\sigma(x) = \\tau(x) \\ne x$ otherwise.\n14. Moreover $\\sigma(i) \\ne z$. Otherwise $i = j$, so $\\tau(i) = i$, which is false for a derangement. Thus $\\sigma \\in E_i''$.\n15. The constructions of steps 9 and 12 are inverse to each other. Starting from $\\sigma$, we get $\\tau(j) = i$, so step 12 recovers the same $j$ and then $\\sigma$. Starting from $\\tau$, we get $\\sigma^{-1}(z) = j$, so step 9 recovers $\\tau$. Hence $|E_i''| = D_{n+1}$.\n16. By the [[rule of sum|0EE2V3CHFK]], $|E_i| = |E_i'| + |E_i''| = D_n + D_{n+1}$.\n17. By step 3, the derangements of $X$ are the disjoint union of the sets $E_i$ for the $n + 1$ elements $i \\in Y$.\n18. By the rule of sum, $D_{n+2} = (n + 1)(D_{n+1} + D_n)$.\n\n## Idea\n\nLook at where the last point $z$ goes, say to $i$. If $\\sigma$ also sends $i$ back to $z$, removing the swap leaves a derangement of $n$ points. Otherwise, short-circuiting $z$ out of $\\sigma$ leaves a derangement of $n + 1$ points.\n",
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        "locator": "Counting derangements",
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