Power Series Expansion for Integer Power of Exponential Function minus 1/Proof

Proof
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:

Thus $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

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:

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: \displaystyle \left({e^z - 1}\right)^n = n! \sum_{k \mathop \in \Z} \left\{ {k \atop n}\right\} \frac {z^k} {k!}$