Stirling Number of the Second Kind of Number with Greater

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

Let $k > n$.

Let $\displaystyle \left\{ {n \atop k}\right\}$ denote a Stirling number of the second kind.

Then:
 * $\displaystyle \left\{ {n \atop k}\right\} = 0$

Proof
By definition, the Stirling numbers of the second kind are defined as the coefficients $\displaystyle \left\{ {n \atop k}\right\}$ which satisfy the equation:


 * $\displaystyle x^n = \sum_k \left\{ {n \atop k}\right\} x^{\underline k}$

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

Both of the expressions on the and  are polynomials in $x$ of degree $n$.

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

Also see

 * Unsigned Stirling Number of the First Kind of Number with Greater equals Zero
 * Signed Stirling Number of the First Kind of Number with Greater equals Zero