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

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

Then:
 * $\map s {n + 1, 1} = \paren {-1}^n n!$

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

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


 * $\ds {n + 1 \brack 1} = \paren {-1}^{n + 1 + 1} \map s {n + 1, 1}$

where $\ds {n + 1 \brack 1}$ denotes an unsigned Stirling number of the first kind.

We have that:
 * $\paren {-1}^{n + 1 + 1} = \paren {-1}^n$

and so:
 * $\ds {n + 1 \brack 1} = \paren {-1}^n \map s {n + 1, 1}$

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