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

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

Then:
 * $s \left({n + 1, 0}\right) = 0$

where $s \left({n + 1, 0}\right)$ denotes an unsigned Stirling number of the first kind.

Proof
By definition of signed Stirling number of the first kind:
 * $s \left({n, k}\right) = \delta_{n k}$

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


 * Particular Values of Signed Stirling Numbers of the First Kind