# Stirling Number of the Second Kind of Number with Self

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## Contents

## Theorem

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

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

## Proof

The proof proceeds by induction.

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

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

### Basis for the Induction

$\map P 0$ is the case:

\(\displaystyle {0 \brace 0}\) | \(=\) | \(\displaystyle \delta_{0 0}\) | 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 $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

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

from which it is to be shown that:

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

### Induction Step

This is the induction step:

\(\displaystyle {k + 1 \brace k + 1}\) | \(=\) | \(\displaystyle \paren {k + 1} {k \brace k + 1} + {k \brace k}\) | Definition of Stirling Numbers of the Second Kind | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {k + 1} \times 0 + {k \brace k}\) | Stirling Number of Number with Greater | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle {k \brace k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | Induction Hypothesis |

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

Therefore:

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

$\blacksquare$

## Also see

- Unsigned Stirling Number of the First Kind of Number with Self
- Signed Stirling Number of the First Kind of Number with Self

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(48)$