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

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

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

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

Proof
By Relation between Signed and Unsigned Stirling Numbers of the First Kind:


 * $\displaystyle \left[{n + 1 \atop 1}\right] = \left({-1}\right)^{n + 1 + 1} s \left({n + 1, 1}\right)$

where $\displaystyle \left[{n + 1 \atop 1}\right]$ denotes an unsigned Stirling number of the first kind.

We have that:
 * $\left({-1}\right)^{n + 1 + 1} = \left({-1}\right)^n$

and so:
 * $\displaystyle \left[{n + 1 \atop 1}\right] = \left({-1}\right)^n s \left({n + 1, 1}\right)$

The result follows from Unsigned Stirling Number of the First Kind of Number with Self.

Also see

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


 * Particular Values of Signed Stirling Numbers of the First Kind