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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 $n\ge1$ (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.

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.

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 $n\ge1$ (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 $S(n,m)$.

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 $n\ge1$ (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.

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 $$ x^n=\sum_{m=0}^n S(n,m) x(x-1)\cdots(x-m+1). $$ Assuming here $n\ge1$ (so that $S(n,0)=0$), dividing both sides by $x$, and then letting $x=0$, we get the desired identity $$\sum_{m=1}^n S(n,m)(-1)^m(m-1)!=0.$$

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 $n\ge1$ (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 $S(n,m)$.