Stirling Number of the Second Kind of n+1 with 0

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

Then:
 * $\displaystyle \left\{ {n + 1 \atop 0}\right\} = 0$

where $\displaystyle \left\{ {n + 1 \atop 0}\right\}$ denotes a Stirling number of the second kind.

Proof
We are given that $k = 0$.

So by definition of unsigned Stirling number of the first kind:
 * $\displaystyle \left\{ {n \atop k}\right\} = \delta_{n k}$

where $\delta_{n k}$ is the Kronecker delta.

Thus

Hence the result.

Also see

 * Unsigned Stirling Number of the First Kind of n+1 with 0
 * Signed Stirling Number of the First Kind of n+1 with 0


 * Particular Values of Stirling Numbers of the Second Kind