Proving a function $f(x + T)=k\;f(x)$ satisfies $f(x)=a^x g(x)$ for periodical $g$ I need to prove the following:

If a function $\,f$ satisfies
$$f(x+T)=k\;f(x), \forall x \in \mathbb R$$
for some $k \in \mathbb N$ and $T > 0$, prove that $\,f$ can be written as $f(x)=a^xg(x)$ where $g$ is a periodical function with period $T$. Prove reverse statement/reversal.

I would need some tips/hints how to begin, since I have no idea how to start.

If you do the reverse statement, then it is suggestive that $a=k^{1/T}$.
So let us set $a=k^{1/T}$ and consider $g(x)\equiv f(x)/a^x$. To check the periodicity of $g$:
$$
g(x+T)=\frac{f(x+T)}{a^{x+T}}=\frac{kf(x)}{a^{x+T}}=\frac{a^Tf(x)}{a^{x+T}}=\frac{f(x)}{a^x}=g(x).
$$
