Considering $k,x\in \mathbb{R}, k^x=1+\sum_{n=1^{\infty\frac{x^n(\ln(k))^n}{n!}$.
Does it have an elementary proof?
I attempt to solve by Maclaurin Series:
$f(x)=\sum_{n=1}^{\infty}\frac{x^n f^{(n)}(0)}{n!}$.
It is well known that the $n$th derivatives of the function $k^x$ are $\frac{d}{dx}k^x=k^x\ln (k)$; $\frac{d^2}{dx^2}k^x=k^x(\ln(k))^2$ and, consequently, $\frac{d^n}{dx^n}k^x=k^x(\ln (k))^n$.
So$k^x=\sum \frac{x^n (k^0)^{(n)}}{n!}=\sum \frac{x^n (k^0 \ln (k))^{(n)}}{n!}=\sum_0^\infty \frac{x^n (\ln (k))^n}{n!}.$
The first term, when $n=0$ is $1$. Therefore, all the summation may be rewritten as $1+\sum_1^\infty \frac{x^n (\ln(k))^n}{n!}$.
Notice $k^x=e^{\ln(k^x)}=e^{x{\ln(k)}}$ by exponent rules, and the taylor series for $e^x$ gives rise to this identity. The 1 term arises from n=0 in the taylor series for $e^x.$ If $k^x$ is complex you can take the principal branch of the logarithm, which will give the same answer for the complex series of $e^x$