# Stirling Number of the Second Kind of Number with Self

## Theorem

$\displaystyle \left\{ {n \atop n}\right\} = 1$

where $\displaystyle \left\{ {n \atop n}\right\}$ denotes a Stirling number of the second kind.

## Proof

The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \left\{ {n \atop n}\right\} = 1$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle \left\{ {0 \atop 0}\right\}$ $=$ $\displaystyle \delta_{0 0}$ Stirling Number of the Second Kind of 0 $\displaystyle$ $=$ $\displaystyle 1$ Definition of Kronecker Delta

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle \left\{ {k \atop k}\right\} = 1$

from which it is to be shown that:

$\displaystyle \left\{ {k + 1 \atop k + 1}\right\} = 1$

### Induction Step

This is the induction step:

 $\displaystyle \left\{ { {k + 1} \atop {k + 1} }\right\}$ $=$ $\displaystyle \left({k + 1}\right) \left\{ {k \atop {k + 1} }\right\} + \left\{ {k \atop k}\right\}$ Definition of Stirling Numbers of the Second Kind $\displaystyle$ $=$ $\displaystyle \left({k + 1}\right) \times 0 + \left\{ {k \atop k}\right\}$ Stirling Number of Number with Greater $\displaystyle$ $=$ $\displaystyle \left\{ {k \atop k}\right\}$ $\displaystyle$ $=$ $\displaystyle 1$ Induction Hypothesis

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

Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: \left\{ {n \atop n}\right\} = 1$

$\blacksquare$