# Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k

## Theorem

Let $m, n \in \Z_{\ge 0}$.

$\displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$

where $\displaystyle {k \brace m}$ etc. denotes a Stirling number of the second kind.

## Proof

The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\displaystyle \forall n \in \Z_{\ge 0}: \displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$

### Basis for the Induction

$\map P 0$ is the case:

 $\displaystyle$  $\displaystyle \sum_{k \mathop \le 0} {k \brace m} \paren {m + 1}^{0 - k}$ $\displaystyle$ $=$ $\displaystyle {0 \brace m} \paren {m + 1}^{0 - 0}$ $\displaystyle$ $=$ $\displaystyle \delta_{0 m}$ Definition of Stirling Numbers of the Second Kind $\displaystyle$ $=$ $\displaystyle \delta_{1 \paren {m + 1} }$ $\displaystyle$ $=$ $\displaystyle {1 \brace {m + 1} }$ Stirling Number of the Second Kind of 1

So $\map P 0$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:

$\displaystyle \sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} = {{r + 1} \brace {m + 1}}$

from which it is to be shown that:

$\displaystyle \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} = { {r + 2} \brace {m + 1} }$

### Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k}$ $\displaystyle$ $=$ $\displaystyle { {r + 1} \brace m} \paren {m + 1}^{r + 1 - r - 1} + \paren {m + 1} \paren {\sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} }$ $\displaystyle$ $=$ $\displaystyle { {r + 1} \brace m} + \paren {m + 1} { {r + 1} \brace {m + 1} }$ $\displaystyle$ $=$ $\displaystyle { {r + 2} \brace {m + 1} }$ Definition of Stirling Numbers of the Second Kind

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$

$\blacksquare$