# Identity involving Stirling number of the second kind

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.

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.
• You can also substitute $x=-1$ Commented Jan 19, 2023 at 18:49
• @მამუკაჯიბლაძე : Thank you for your comment. Then you will probably need some other identity as well, because on the right-hand side you will have $\sum_{m=1}^n S(n,m)(-1)^m m!$ rather than $\sum_{m=1}^n S(n,m)(-1)^m(m-1)!$ (also $(-1)^n$ instead of $0$ on the left). Commented Jan 19, 2023 at 18:59