# Unsigned Stirling Number of the First Kind of Number with Greater

## Theorem

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

Let $k > n$.

Let $\displaystyle {n \brack k}$ denote an unsigned Stirling number of the first kind.

Then:

$\displaystyle {n \brack k} = 0$

## Proof 1

By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\displaystyle \left[{n \atop k}\right]$ which satisfy the equation:

$\displaystyle x^{\underline n} = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k$

where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.

Both of the expressions on the left hand side and right hand side are polynomials in $x$ of degree $n$.

Hence the coefficient $\displaystyle \left[{n \atop k}\right]$ of $x^k$ where $k > n$ is $0$.

$\blacksquare$

## Proof 2

The proof proceeds by induction.

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

$\displaystyle k > n \implies \left[{n \atop k}\right] = 0$

### Basis for the Induction

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

$\displaystyle \left[{0 \atop k}\right] = \delta_{0 k}$

So by definition of Kronecker delta:

$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle \left[{0 \atop k}\right] = 0$

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

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

$\displaystyle k > r \implies \left[{r \atop k}\right] = 0$

from which it is to be shown that:

$\displaystyle k > r + 1 \implies \left[{r + 1 \atop k}\right] = 0$

### Induction Step

This is the induction step:

 $\displaystyle \left[{r + 1 \atop k}\right]$ $=$ $\displaystyle r \left[{r \atop k}\right] + \left[{r \atop k - 1}\right]$ $\displaystyle$ $=$ $\displaystyle r \times 0 + \left[{r \atop k - 1}\right]$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle r \times 0 + 0$ Induction Hypothesis: $k > r + 1 \implies k - 1 > r$ $\displaystyle$ $=$ $\displaystyle 0$

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

Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies \left[{n \atop k}\right] = 0$

$\blacksquare$