# Stirling Number of the Second Kind of n+1 with 1

## Theorem

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

Then:

$\displaystyle {n + 1 \brace 1} = 1$

where $\displaystyle {n + 1 \brace 1}$ denotes a Stirling number of the second kind.

## Proof

The proof proceeds by induction.

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

$\displaystyle {n + 1 \brace 1} = 1$

### Basis for the Induction

$\map P 0$ is the case:

 $\ds {1 \brace 1}$ $=$ $\ds \delta_{1 1}$ Stirling Number of the Second Kind of 1 $\ds$ $=$ $\ds 1$ Definition of Kronecker Delta

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\displaystyle {k + 1 \brace 1} = 1$

from which it is to be shown that:

$\displaystyle {k + 2 \brace 1} = 1$

### Induction Step

This is the induction step:

 $\ds {k + 2 \brace 1}$ $=$ $\ds 1 \times {k + 1 \brace 1} + {k + 2 \brace 0}$ Definition of Stirling Numbers of the Second Kind $\ds$ $=$ $\ds {k + 1 \brace 1} + 0$ Stirling Number of the Second Kind of n+1 with 0 $\ds$ $=$ $\ds 1$ Induction Hypothesis

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

Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: {n + 1 \brace 1} = 1$

$\blacksquare$