Unsigned Stirling Number of the First 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 an unsigned Stirling number of the first 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

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


 * Particular Values of Unsigned Stirling Numbers of the First Kind