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

## Theorem

Let $e^z$ denote the exponential function.

Then:

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

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:

 $\ds 0! \sum_{k \mathop \in \Z} \left\{ {k \atop 0}\right\} \frac {z^k} {k!}$ $=$ $\ds \sum_{k \mathop \in \Z} \left\{ {k \atop 0}\right\} \frac {z^k} {k!}$ 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 $\ds$ $=$ $\ds \sum_{k \mathop = 0} \frac {z^k} {k!}$ $\ds$ $=$ $\ds \frac {z^0} {0!}$ $\ds$ $=$ $\ds 1$ $\ds$ $=$ $\ds \left({e^z - 1}\right)^0$

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

### Basis for the Induction

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

 $\ds 1! \sum_{k \mathop \in \Z} \left\{ {k \atop 1}\right\} \frac {z^k} {k!}$ $=$ $\ds \sum_{k \mathop \in \Z} \left\{ {k \atop 1}\right\} \frac {z^k} {k!}$ 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 $\ds$ $=$ $\ds \left({e^z - 1}\right)^1$

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:

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

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

$\blacksquare$