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 $\map P n$ be the proposition:
 * $\ds \paren {e^z - 1}^n = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}$

$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \paren {e^z - 1}^r = r! \sum_{k \mathop \in \Z} {k \brace r} \frac {z^k} {k!}$

from which it is to be shown that:
 * $\ds \paren {e^z - 1}^{r + 1} = \paren {r + 1}! \sum_{k \mathop \in \Z} {k \brace r + 1} \frac {z^k} {k!}$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: \ds \paren {e^z - 1}^n = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}$