The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left({e^z - 1}\right)^n = n! \sum_{k \mathop \in \Z} \left\{ {k \atop n}\right\} \frac {z^k} {k!}$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = 0! \sum_{k \mathop \in \Z} \left\{ {k \atop 0}\right\} \frac {z^k} {k!}       | r = \sum_{k \mathop \in \Z} \left\{ {k \atop 0}\right\} \frac {z^k} {k!}       | c = {{Defof|Factorial}}: $0! = 1$ }} {{eqn | r = \sum_{k \mathop \in \Z} \delta_{k 0} \frac {z^k} {k!}       | c = {{Defof|Stirling Numbers of the Second Kind|index = 1}} }} {{eqn | r = \sum_{k \mathop = 0} \frac {z^k} {k!}       | c =  }} {{eqn | r = \frac {z^0} {0!}       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = \left({e^z - 1}\right)^0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = 1! \sum_{k \mathop \in \Z} \left\{ {k \atop 1}\right\} \frac {z^k} {k!}       | r = \sum_{k \mathop \in \Z} \left\{ {k \atop 1}\right\} \frac {z^k} {k!}       | c = {{Defof|Factorial}}: $1! = 1$ }} {{eqn | r = \sum_{k \mathop \ge 1} \frac {z^k} {k!}       | c = Stirling Number of the Second Kind of n+1 with 1 }} {{eqn | r = \sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1       | c =  }} {{eqn | r = e^z - 1       | c = Power Series Expansion for Exponential Function }} {{eqn | r = \left({e^z - 1}\right)^1       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left({e^z - 1}\right)^r = r! \sum_{k \mathop \in \Z} \left\{ {k \atop r}\right\} \frac {z^k} {k!}$ from which it is to be shown that: :$\displaystyle \left({e^z - 1}\right)^{r + 1} = \left({r + 1}\right)! \sum_{k \mathop \in \Z} \left\{ {k \atop {r + 1} }\right\} \frac {z^k} {k!}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left({e^z - 1}\right)^{r + 1}       | r = \left({e^z - 1}\right)^r \left({e^z - 1}\right)       | c =  }} {{eqn | r = \left({r! \sum_{k \mathop \in \Z} \left\{ {k \atop r}\right\} \frac {z^k} {k!} }\right) \left({e^z - 1}\right)       | c = Induction Hypothesis }} {{eqn | r = \left({r! \sum_{k \mathop \in \Z} \left\{ {k \atop r}\right\} \frac {z^k} {k!} }\right) \left({\sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1}\right)       | c = Power Series Expansion for Exponential Function }} {{eqn | r = \left({r! \sum_{k \mathop \ge 0} \left\{ {k \atop r}\right\} \frac {z^k} {k!} }\right) \left({\sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1}\right)       | c = $\displaystyle \left\{ {k \atop r}\right\} = 0$ for $k < 0$ }} {{eqn | r = r! \sum_{k \mathop \ge 0} \binom r k \left\{ {k \atop r}\right\} \frac {z^k} {k!} - r! \sum_{k \mathop \ge 0} \left\{ {k \atop r}\right\} \frac {z^k} {k!}       | c = Product of Exponential Generating Functions }} {{eqn | r = r! \sum_{k \mathop \ge 0} \left\{ { {k + 1} \atop {r + 1} }\right\} \frac {z^k} {k!} - r! \sum_{k \mathop \ge 0} \left\{ {k \atop r}\right\} \frac {z^k} {k!}       | c = Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k }} {{eqn | r = r! \sum_{k \mathop \ge 0} \left({\left({r + 1}\right) \left\{ {k \atop {r + 1} }\right\} + \left\{ {k \atop r}\right\} - \left\{ {k \atop r}\right\} }\right) \frac {z^k} {k!}       | c = {{Defof|Stirling Numbers of the Second Kind|index = 1}} }} {{eqn | r = \left({r + 1}\right)! \sum_{k \mathop \ge 0} \left\{ {k \atop {r + 1} }\right\} \frac {z^k} {k!}       | c =  }} {{eqn | r = \left({r + 1}\right)! \sum_{k \mathop \in \Z} \left\{ {k \atop {r + 1} }\right\} \frac {z^k} {k!}       | c = $\displaystyle \left\{ {k \atop r}\right\} = 0$ for $k < 0$ }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \displaystyle \left({e^z - 1}\right)^n = n! \sum_{k \mathop \in \Z} \left\{ {k \atop n}\right\} \frac {z^k} {k!}$ {{qed}}