Power Series Expansion for Integer Power of Exponential Function minus 1

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Theorem

Let $e^z$ denote the exponential function.


Then:

\(\displaystyle \left({e^z - 1}\right)^n\) \(=\) \(\displaystyle z^n + \dfrac 1 {n + 1} \left\{ { {n + 1} \atop n}\right\} z^{n + 1} + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle n! \sum_{k \mathop \in \Z} \left\{ {k \atop n}\right\} \frac {z^k} {k!}\)

where $\displaystyle \left\{ {k \atop n}\right\}$ denotes a Stirling number of the second kind.


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:

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

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


Basis for the Induction

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

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

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:

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


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


Therefore:

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

$\blacksquare$


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