Wikipedia has the identity $$S(n+1, k+1) = \sum_i {n \choose i} S(i, k)\;.$$ I can't seem to find a proof for this and was wondering what it entails. Is it a purely combinatoric argument, or an algebraic manipulation of the explicit formula $S(n, k) = \frac{1}{k!} \sum_{i=0}^k (-1)^i {k \choose i}(k-i)^n $ with the known recurrence formula $S(n+1, k+1) = S(n, k) + (k+1)S(n, k+1)? $
Proof of recursion formula for Stirling number of the second kind
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combinatorics
discrete-mathematics
recurrence-relations
stirling-numbers
1 Answers
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$S(n+1,k+1)$ is the number of ways for partitioning a set with $n+1$ elements in $k+1$ disjoint, non-empty subsets. Assume that $A_1\cup A_2\cup\ldots\cup A_{k+1}$ is a partition of $\{1,\ldots,n+1\}$. We may count such partitions according to the size of $A_{k+1}$ and its elements. If we assume that $\left|A_{k+1}\right|=n-i$, there are $\binom{n}{n-i}=\binom{n}{i}$ for choosing the elements of $A_{k+1}$ and $S(i,k)$ ways for partitioning $\{1,\ldots,n+1\}\setminus A_{k+1}$ in $k$ pieces.
$$ S(n+1,k+1) = \sum_{i=0}^{n-1}\binom{n}{i} S(i,k) $$ immediately follows.