Proof $(e^x-1)^r = \sum_{n=0}^{\infty} \sum_{k=0}^r (-1)^k \binom{r}{k} (r-k)^n \frac{x^n}{n!}$

Question

In our combinatorics script I found this

$$(e^x-1)^r = \sum_{n=0}^{\infty} \sum_{k=0}^r (-1)^k \binom{r}{k} (r-k)^n \frac{x^n}{n!}$$

I tried finding a proof for this on the stackexchange math and google, but I couldn't.

How can one prove this?

Answer 1

$$\begin{align*} (e^x - 1)^r & = \sum_{k = 0}^r \binom{r}{k} (-1)^k (e^x)^{r-k} \\ & = \sum_{k = 0}^r \binom{r}{k} (-1)^k e^{x(r-k)} \\ & = \sum_{k = 0}^r \binom{r}{k} (-1)^k \sum_{n = 0}^\infty \frac{(x(r-k))^n}{n!} \\ & = \sum_{n = 0}^\infty \sum_{k = 0}^r (-1)^k \binom{r}{k} (r-k)^n \frac{x^n}{n!} \end{align*}$$ where we used the binomial theorem in the first line and Taylor series expansion for $e^x$ in the last line.