Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial

Theorem

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

$\displaystyle \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$

where:

$\displaystyle {k \brack m}$ denotes an unsigned Stirling number of the first kind
$n!$ denotes a factorial.

Proof

The proof proceeds by induction on $n$.

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

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$

Basis for the Induction

$\map P 0$ is the case:

 $\displaystyle$  $\displaystyle \sum_{k \mathop \le 0} {k \brack m} \frac {0!} {k!}$ $\displaystyle$ $=$ $\displaystyle {0 \brack m} \frac 1 {0!}$ Definition of Unsigned Stirling Numbers of the First Kind $\displaystyle$ $=$ $\displaystyle \delta_{0 m}$ Unsigned Stirling Number of the First Kind of 0 $\displaystyle$ $=$ $\displaystyle \delta_{1 \paren {m + 1} }$ $\displaystyle$ $=$ $\displaystyle {1 \brack m + 1}$ Unsigned Stirling Number of the First 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:

$\displaystyle \sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} = {r + 1 \brack m + 1}$

from which it is to be shown that:

$\displaystyle \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!} = {r + 2 \brack m + 1}$

Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!}$ $\displaystyle$ $=$ $\displaystyle {r + 1 \brack m} \frac {\paren {r + 1}!} {\paren {r + 1}!} + \paren {r + 1} \paren {\sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} }$ $\displaystyle$ $=$ $\displaystyle {r + 1 \brack m} + \paren {r + 1} {r + 1 \brack m + 1}$ $\displaystyle$ $=$ $\displaystyle {r + 2 \brack m + 1}$ Definition of Unsigned Stirling Numbers of the First Kind

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

Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$

$\blacksquare$