I defined $a^z$ for $z \in \mathbb{C}$ as
$a^z = \exp(z\log(a))$
and I proved it is continous, now I want to show that
$a^n = a \cdot a \cdot a \cdot \ldots \cdot a$ for $n \in \mathbb{N}$
so since $n$ is just a real number here I thought this wouldnt be to bad, but I'm stuck.
$a^n = \large\exp(n\log(a)) = \sum\limits_{k=0}^\infty \frac{(n\log(a))^k}{k!} = \frac{n^k*\log(a)^k}{k!} = \frac{k(n^k)\log(a)}{k!}$
I don't know where to go from here though.
\begin{eqnarray} a^n &=& \exp(n\log(a)), \\ &=& \exp(\log(a) + \ldots + \log(a)), \quad \mbox{n times}\\ &=& \exp(\log(a))\exp(\log(a))\ldots \exp(\log(a)) \\ &=& a a \ldots a. \end{eqnarray}