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.

$\blacksquare$

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