Unsigned Stirling Number of the First Kind of 0

Theorem

 * $\ds {0 \brack n} = \delta_{0 n}$

where:
 * $\ds {0 \brack n}$ denotes an unsigned Stirling number of the first kind
 * $\delta_{0 n}$ denotes the Kronecker delta.

Proof
By definition of unsigned Stirling number of the first kind:

$\ds x^{\underline 0} = \sum_k \paren {-1}^{0 - k} {0 \brack k} x^k$

Thus we have:

Thus, in the expression:
 * $\ds x^{\underline 0} = \sum_k \paren {-1}^{-k} {0 \brack k} x^k$

we have:
 * $\ds {0 \brack 0} = 1$

and for all $k \in \Z_{>0}$:
 * $\ds {0 \brack k} = 0$

That is:
 * $\ds {0 \brack k} = \delta_{0 k}$

Also see

 * Signed Stirling Number of the First Kind of 0
 * Stirling Number of the Second Kind of 0


 * Particular Values of Unsigned Stirling Numbers of the First Kind