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 $\ds {n \brack n - m}$ 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 $\map P n$ be the proposition:
 * $\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.

Basis for the Induction
$\map P 0$ is the case:
 * $\ds {n \brack n} = 1$

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

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds {n \brack n - k}$ is a polynomial in $n$ of degree $2 k$.

from which it is to be shown that:
 * $\ds {n \brack n - \paren {k + 1} }$ is a polynomial in $n$ of degree $2 \paren {k + 1}$.

Induction Step
This is the induction step:

Let $\map f {n, d}$ denote an arbitrary polynomial in $n$ of degree $d$.

Then:

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

Therefore:
 * $\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.