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  "body": "\n## Proof\n\nWe sort the partitions by the elements outside the block of the last element.\n\n1. Let $n$ be a natural number. Let $X = \\set{1, \\dots, n + 1}$, let $z = n + 1$, and let $Y = \\set{1, \\dots, n}$.\n2. By the definition of the [[Bell numbers|0WW85SQVS3]], $B_m$ is the number of [[partitions|09CCE86A6D]] of an $m$-element set. So $B_{n+1}$ counts the partitions of $X$.\n3. Let $P$ be a partition of $X$, and let $B$ be its block that contains $z$. Let $K = X \\setminus B$. Then $K \\subseteq Y$.\n4. The other blocks $P \\setminus \\set{B}$ are nonempty, pairwise disjoint, and their union is $K$. So they form a partition of $K$.\n5. Map $P$ to the pair $(K, P \\setminus \\set{B})$.\n6. Conversely, let $K \\subseteq Y$ and let $Q$ be a partition of $K$. The set $X \\setminus K$ contains $z$, so it is nonempty. Then $Q \\cup \\set{X \\setminus K}$ is a partition of $X$.\n7. These two maps are inverse to each other. So the partitions of $X$ are in bijection with the pairs $(K, Q)$, where $K \\subseteq Y$ and $Q$ is a partition of $K$.\n8. For a fixed $K$ with $k$ elements, there are $B_k$ choices of $Q$.\n9. By the [[number of $k$-element subsets|0BFMR2FQ8B]], $Y$ has $\\binom{n}{k}$ subsets $K$ with $k$ elements, for $0 \\le k \\le n$.\n10. By the [[rule of sum|0EE2V3CHFK]], grouping the pairs by $k = |K|$, and by the [[bijection principle|0GV2573FP4]],\n$$B_{n+1} = \\sum_{k=0}^{n} \\binom{n}{k} B_k.$$\n\n## Idea\n\nIn a partition of $\\set{1, \\dots, n + 1}$, the block of the last element leaves out some set $K$ of the first $n$ elements. Choose $K$ with $k$ elements in $\\binom{n}{k}$ ways, then partition $K$ freely in $B_k$ ways.\n",
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