Theorem: Power Series Expansion for Integer Power of Exponential Function minus 1

Let $e^z$ denote the exponential function.
Then:
{{begin-eqn}}
{{eqn | l = \paren {e^z - 1}^n
      | r = z^n + \dfrac 1 {n + 1} {n + 1 \brace n} z^{n + 1} + \cdots
      | c = 
}}
{{eqn | r = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}
      | c = 
}}
{{end-eqn}}
where $\ds {k \brace n}$ denotes a Stirling number of the second kind.

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:
{{begin-eqn}}
{{eqn | l = 0! \sum_{k \mathop \in \Z} {k \brace 0} \frac {z^k} {k!}
      | r = \sum_{k \mathop \in \Z} {k \brace 0} \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 = \paren {e^z - 1}^0
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = 1! \sum_{k \mathop \in \Z} {k \brace 1} \frac {z^k} {k!}
      | r = \sum_{k \mathop \in \Z} {k \brace 1} \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 = \paren {e^z - 1}^1
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \paren {e^z - 1}^{r + 1}
      | r = \paren {e^z - 1}^r \paren {e^z - 1}
      | c = 
}}
{{eqn | r = \paren {r! \sum_{k \mathop \in \Z} {k \brace r} \frac {z^k} {k!} } \paren {e^z - 1}
      | c = Induction Hypothesis
}}
{{eqn | r = \paren {r! \sum_{k \mathop \in \Z} {k \brace r} \frac {z^k} {k!} } \paren {\sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1}
      | c = Power Series Expansion for Exponential Function
}}
{{eqn | r = \paren {r! \sum_{k \mathop \ge 0} {k \brace r} \frac {z^k} {k!} } \paren {\sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1}
      | c = $\ds {k \brace r} = 0$ for $k < 0$
}}
{{eqn | r = r! \sum_{k \mathop \ge 0} \binom r k {k \brace r} \frac {z^k} {k!} - r! \sum_{k \mathop \ge 0} {k \brace r} \frac {z^k} {k!}
      | c = Product of Exponential Generating Functions
}}
{{eqn | r = r! \sum_{k \mathop \ge 0} { {k + 1} \brace r + 1} \frac {z^k} {k!} - r! \sum_{k \mathop \ge 0} {k \brace r} \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} \paren {\paren {r + 1} {k \brace r + 1} + {k \brace r} - {k \brace r} } \frac {z^k} {k!}
      | c = {{Defof|Stirling Numbers of the Second Kind|index = 1}}
}}
{{eqn | r = \paren {r + 1}! \sum_{k \mathop \ge 0} {k \brace r + 1} \frac {z^k} {k!}
      | c = 
}}
{{eqn | r = \paren {r + 1}! \sum_{k \mathop \in \Z} {k \brace r + 1} \frac {z^k} {k!}
      | c = $\ds {k \brace r} = 0$ for $k < 0$
}}
{{end-eqn}}
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \ds \paren {e^z - 1}^n = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}$
{{qed}}