# Summation over Lower Index of Unsigned Stirling Numbers of the First Kind

## Contents

## Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

- $\displaystyle \sum_k \left[{n \atop k}\right] = n!$

where:

- $\displaystyle \left[{n \atop k}\right]$ denotes an unsigned Stirling number of the first kind
- $n!$ denotes the factorial of $n$.

## Proof

The proof proceeds by induction on $n$.

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

- $\displaystyle \sum_k \left[{n \atop k}\right] = n!$

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

\(\displaystyle \sum_k \left[{0 \atop k}\right]\) | \(=\) | \(\displaystyle \sum_k \delta_{0 k}\) | Unsigned Stirling Number of the First Kind of 0 | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | all terms vanish but for $k = 0$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0!\) | Definition of Factorial |

Thus $P \left({0}\right)$ is seen to hold.

### Basis for the Induction

$P \left({1}\right)$ is the case:

\(\displaystyle \sum_k \left[{1 \atop k}\right]\) | \(=\) | \(\displaystyle \sum_k \delta_{1 k}\) | Unsigned Stirling Number of the First Kind of 1 | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | all terms vanish but for $k = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1!\) | Definition of Factorial |

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

- $\displaystyle \sum_k \left[{m \atop k}\right] = m!$

from which it is to be shown that:

- $\displaystyle \sum_k \left[{m + 1 \atop k}\right] = \left({m + 1}\right)!$

### Induction Step

This is the induction step:

\(\displaystyle \sum_k \left[{m + 1 \atop k}\right]\) | \(=\) | \(\displaystyle \sum_k \left({m \left[{m \atop k}\right] + \left[{m \atop k - 1}\right]}\right)\) | Definition of Unsigned Stirling Numbers of the First Kind | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m \sum_k \left[{m \atop k}\right] + \sum_k \left[{m \atop k - 1}\right]\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m \sum_k \left[{m \atop k}\right] + \sum_k \left[{m \atop k}\right]\) | Translation of Index Variable of Summation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + 1}\right) \sum_k \left[{m \atop k}\right]\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + 1}\right) m!\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + 1}\right)!\) | Definition of Factorial |

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

Therefore:

- $\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] = n!$

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $39$