# Stirling Number of n with n-m is Polynomial in n of Degree 2m/Unsigned First Kind

## Theorem

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

The unsigned Stirling number of the first kind $\displaystyle \left[{n \atop n - m}\right]$ is a polynomial in $n$ of degree $2 m$.

## Proof

The proof proceeds by induction over $m$.

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

$\displaystyle \left[{n \atop n - m}\right]$ is a polynomial in $n$ of degree $2 m$.

### Basis for the Induction

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

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

which is a polynomial in $n$ of degree $0$

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

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

$\displaystyle \left[{n \atop n - k}\right]$ is a polynomial in $n$ of degree $2 k$.

from which it is to be shown that:

$\displaystyle \left[{n \atop n - \left({k + 1}\right)}\right]$ is a polynomial in $n$ of degree $2 \left({k + 1}\right)$.

### Induction Step

This is the induction step:

Let $f \left({n, d}\right)$ denote an arbitrary polynomial in $n$ of degree $d$.

Then:

 $\displaystyle \left[{n \atop n - \left({k + 1}\right)}\right]$ $=$ $\displaystyle \left[{n \atop n - k - 1}\right]$ $\displaystyle$ $=$ $\displaystyle \left({n - 1}\right) \left[{n - 1 \atop n - k - 1}\right] + \left[{n - 1 \atop n - k - 2}\right]$ Definition 1 of Unsigned Stirling Number of the First Kind $\displaystyle$ $=$ $\displaystyle \left({n - 1}\right) \left[{n - 1 \atop \left({n - 1}\right) - k}\right] + \left[{n - 1 \atop \left({n - 1}\right) - \left({k + 1}\right)}\right]$

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

Therefore:

$\displaystyle \left[{n \atop n - m}\right]$ is a polynomial in $n$ of degree $2 m$.