Identity involving Stirling number of the second kind

Question

I'm looking for a citable reference for the following identity involving the Stirling numbers of the second kind $S(n, k)$ stated in Equation (27): For $n \geq 2$, $$ \sum_{m=1}^n S(n, m) (-1)^m (m-1)! = 0. $$ Thank you.

Answer 1

The first formula in Section 24.1.4.I.B in Handbook of mathematical functions with formulas, graphs, and mathematical tables, Dover edition, 1965 (Library of Congress Catalog Card Number: 65.12253) by Abramowitz and Stegun is $$ x^n=\sum_{m=0}^n S(n,m) x(x-1)\cdots(x-m+1). \tag{1}\label{1} $$ Assuming here $n>1$ (so that $S(n,0)=0$), dividing both sides by $x$, and then letting $x=0$, we get the desired identity $$\sum_{m=1}^n S(n,m)(-1)^m(m-1)!=0.$$

---

One may note that Riordan (An Introduction to Combinatorial Analysis, Wiley, 1958, formula (35) on p. 33) uses identity \eqref{1} to define the Stirling numbers $S(n,m)$, then deriving their combinatorial definition (as the number of ways of partitioning a set of $n$ elements into $m$ nonempty sets) -- cf. formula (38) on p. 33 and the third display from the bottom on p. 91 of the book.