I have two things to proof.
$(e^{ax})^{-1}=e^{-ax}$ and $(e^{ax})^{m}=e^{(ma)x}$
I know the power series of $e^x$ and $e^{ax}$ and that $e^{ax}e^{bx}=e^{(a+b)x}$
I tried it forward and backward: $(e^{ax})^{-1}=(\sum_{n=0}^{\infty}\frac{1}{n!}(ax)^n)^{-1} = \frac{1}{\sum_{n=0}^{\infty}\frac{1}{n!}(ax)^n}$
and i dont know how to continue. I know that i want this $\sum_{n=0}^{\infty}\frac{1}{n!}(-ax)^n$ or
$\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}(ax)^n$
I think if i know how to solve the first problem, i know how to solve the secound one. They look similar.