Signed Stirling Number of the First Kind of Number with Self

Theorem

 * $s \left({n, n}\right) = 1$

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

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


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

We have that:
 * $\left({-1}\right)^{n + n} = \left({-1}\right)^{2 n} = 1$

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

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 Number with Self
 * Stirling Number of the Second Kind of Number with Self


 * Particular Values of Signed Stirling Numbers of the First Kind