Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1

Theorem

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

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

Then:

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

Proof

By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\ds {n \brack k}$ which satisfy the equation:

$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$

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 $\ds {n \brack k}$ of $x^k$ where $k > n$ is $0$.

$\blacksquare$