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:

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 $2 m$.