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

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Theorem

Let $e^z$ denote the exponential function.


Then:

\(\ds \paren {e^z - 1}^n\) \(=\) \(\ds z^n + \dfrac 1 {n + 1} {n + 1 \brace n} z^{n + 1} + \cdots\)
\(\ds \) \(=\) \(\ds n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}\)

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:

\(\ds 0! \sum_{k \mathop \in \Z} {k \brace 0} \frac {z^k} {k!}\) \(=\) \(\ds \sum_{k \mathop \in \Z} {k \brace 0} \frac {z^k} {k!}\) Definition of Factorial: $0! = 1$
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \Z} \delta_{k 0} \frac {z^k} {k!}\) Definition 1 of Stirling Numbers of the Second Kind
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0} \frac {z^k} {k!}\)
\(\ds \) \(=\) \(\ds \frac {z^0} {0!}\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \paren {e^z - 1}^0\)

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


Basis for the Induction

$\map P 1$ is the case:

\(\ds 1! \sum_{k \mathop \in \Z} {k \brace 1} \frac {z^k} {k!}\) \(=\) \(\ds \sum_{k \mathop \in \Z} {k \brace 1} \frac {z^k} {k!}\) Definition of Factorial: $1! = 1$
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 1} \frac {z^k} {k!}\) Stirling Number of the Second Kind of n+1 with 1
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1\)
\(\ds \) \(=\) \(\ds e^z - 1\) Power Series Expansion for Exponential Function
\(\ds \) \(=\) \(\ds \paren {e^z - 1}^1\)

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:

\(\ds \paren {e^z - 1}^{r + 1}\) \(=\) \(\ds \paren {e^z - 1}^r \paren {e^z - 1}\)
\(\ds \) \(=\) \(\ds \paren {r! \sum_{k \mathop \in \Z} {k \brace r} \frac {z^k} {k!} } \paren {e^z - 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {r! \sum_{k \mathop \in \Z} {k \brace r} \frac {z^k} {k!} } \paren {\sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1}\) Power Series Expansion for Exponential Function
\(\ds \) \(=\) \(\ds \paren {r! \sum_{k \mathop \ge 0} {k \brace r} \frac {z^k} {k!} } \paren {\sum_{k \mathop \ge 0} \frac {z^k} {k!} - 1}\) $\ds {k \brace r} = 0$ for $k < 0$
\(\ds \) \(=\) \(\ds r! \sum_{k \mathop \ge 0} \binom r k {k \brace r} \frac {z^k} {k!} - r! \sum_{k \mathop \ge 0} {k \brace r} \frac {z^k} {k!}\) Product of Exponential Generating Functions
\(\ds \) \(=\) \(\ds r! \sum_{k \mathop \ge 0} { {k + 1} \brace r + 1} \frac {z^k} {k!} - r! \sum_{k \mathop \ge 0} {k \brace r} \frac {z^k} {k!}\) Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k
\(\ds \) \(=\) \(\ds r! \sum_{k \mathop \ge 0} \paren {\paren {r + 1} {k \brace r + 1} + {k \brace r} - {k \brace r} } \frac {z^k} {k!}\) Definition 1 of Stirling Numbers of the Second Kind
\(\ds \) \(=\) \(\ds \paren {r + 1}! \sum_{k \mathop \ge 0} {k \brace r + 1} \frac {z^k} {k!}\)
\(\ds \) \(=\) \(\ds \paren {r + 1}! \sum_{k \mathop \in \Z} {k \brace r + 1} \frac {z^k} {k!}\) $\ds {k \brace r} = 0$ for $k < 0$


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


Therefore:

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

$\blacksquare$


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