Stirling Number of the Second Kind of Number with Greater/Proof 1

Theorem

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

Let $\ds {n \brace k}$ denote a Stirling number of the second kind.

Then:

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

Proof

By definition, the Stirling numbers of the second kind are defined as the coefficients $\ds {n \brace k}$ which satisfy the equation:

$\ds x^n = \sum_k {n \brace k} x^{\underline k}$

where $x^{\underline k}$ denotes the $k$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 $\ds {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.

$\blacksquare$