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 $n \geq 2$, $$ \sum_{m=1}^n S(n, m) (1)^m (m1)! = 0. $$ Thank you.

3$\begingroup$ I would think that MO or MSE are "citable references". There are many, many proofs: math.stackexchange.com/q/1279874/87355 and math.stackexchange.com/questions/395139/… and math.stackexchange.com/q/1125097/87355 and math.stackexchange.com/q/3255200/87355 $\endgroup$– Carlo BeenakkerCommented Jan 19, 2023 at 16:50

1$\begingroup$ You want to require $n \geq 2$. $\endgroup$– darij grinbergCommented Jan 19, 2023 at 17:49

$\begingroup$ This is Problem 3.13 (c) in Ioan Tomescu, Problems in Combinatorics and Graph Theory, Wiley 1985. $\endgroup$– darij grinbergCommented Jan 19, 2023 at 17:50
1 Answer
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 $$ x^n=\sum_{m=0}^n S(n,m) x(x1)\cdots(xm+1). \tag{1}\label{1} $$ 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 $$\sum_{m=1}^n S(n,m)(1)^m(m1)!=0.$$
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.

$\begingroup$ You can also substitute $x=1$ $\endgroup$ Commented Jan 19, 2023 at 18:49

1$\begingroup$ @მამუკაჯიბლაძე : Thank you for your comment. Then you will probably need some other identity as well, because on the righthand side you will have $\sum_{m=1}^n S(n,m)(1)^m m!$ rather than $\sum_{m=1}^n S(n,m)(1)^m(m1)!$ (also $(1)^n$ instead of $0$ on the left). $\endgroup$ Commented Jan 19, 2023 at 18:59

$\begingroup$ Sorry for confusion, indeed this is a different identity! $\endgroup$ Commented Jan 19, 2023 at 19:15