# Sum over k of Stirling Numbers of Second Kind by x^k

## Theorem

 $\displaystyle \sum_k \left\{ {k \atop n}\right\} z^k$ $=$ $\displaystyle \dfrac {z^n} {\prod \limits_{k \mathop = 1}^n \left({1 - k n}\right)}$ $\displaystyle$ $=$ $\displaystyle \dfrac {z^n} {\left({1 - z}\right) \left({1 - 2 z}\right) \cdots \left({1 - n z}\right)}$

where:

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