Stirling Number of the Second Kind of Number with Greater

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

Let $k > n$.

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

Then:
 * $\ds {n \brace k} = 0$

Also see

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