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  "body": "\n## Proof\n\nWe sort the partitions by the block of the last element.\n\n1. Let $n$ and $k$ be natural numbers. Let $X = \\set{1, \\dots, n + 1}$, let $z = n + 1$, and let $Y = \\set{1, \\dots, n}$.\n2. By the definition of the [[Stirling numbers of the second kind|0ZN07RQ3JS]], $S(n + 1, k + 1)$ is the number of [[partitions|09CCE86A6D]] of $X$ into $k + 1$ blocks.\n3. Let $P$ be such a partition, and let $B$ be its block that contains $z$. Either $B = \\set{z}$, or $B$ has another element.\n4. Case $B = \\set{z}$. Then $P \\setminus \\set{B}$ is a partition of $Y$ into $k$ blocks.\n5. Conversely, if $Q$ is a partition of $Y$ into $k$ blocks, then $Q \\cup \\set{\\set{z}}$ is a partition of $X$ into $k + 1$ blocks. One of its blocks is $\\set{z}$.\n6. These maps are inverse to each other. So by the [[bijection principle|0GV2573FP4]], this case has $S(n, k)$ partitions.\n7. Case $B \\ne \\set{z}$. Then $B \\setminus \\set{z}$ is nonempty. Replacing $B$ by $B \\setminus \\set{z}$ gives a partition $P'$ of $Y$ into $k + 1$ blocks.\n8. Map $P$ to the pair $(P', B \\setminus \\set{z})$. The second entry is a block of $P'$.\n9. Conversely, let $Q$ be a partition of $Y$ into $k + 1$ blocks, and let $C \\in Q$. Replace $C$ by $C \\cup \\set{z}$. This gives a partition of $X$ into $k + 1$ blocks. In it, the block of $z$ is not $\\set{z}$.\n10. These maps are inverse to each other. So this case is in bijection with the pairs $(Q, C)$, where $Q$ is a partition of $Y$ into $k + 1$ blocks and $C \\in Q$.\n11. Each such $Q$ has exactly $k + 1$ choices of $C$. By the [[rule of sum|0EE2V3CHFK]] over $Q$, there are $(k + 1)\\, S(n, k + 1)$ such pairs.\n12. The two cases are disjoint and cover all partitions of $X$ into $k + 1$ blocks. By the rule of sum,\n$$S(n + 1, k + 1) = (k + 1)\\, S(n, k + 1) + S(n, k).$$\n\n## Idea\n\nRemove the last element $z$ from its block. If $z$ was alone, one block disappears and $k$ blocks remain. If not, $k + 1$ blocks remain, and $z$ could have been in any one of them, which gives the factor $k + 1$.\n",
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