Show that for $x \in (-\pi, \pi)$ and $T(x) := \sum_{n=0}^\infty(\frac{x}{\pi})^n$ the derivation $T'(x) = \frac{\pi}{(x-\pi)^2}$ $\frac{x}{\pi} < 1$ for $x\in (-\pi, \pi)$, therefore we could use the formula for geometric series to get the limit.
$$\sum_{n = 0}^\infty(\frac{x}{n})^n = \frac{1}{1-\frac{x}{\pi}} = \frac{1}{\frac{\pi}{\pi}-\frac{x}{\pi}} = \frac{1}{\frac{\pi-x}{\pi}} = \frac{\pi}{\pi-x}$$
Now, this doesn't get me any further and im stuck here, so hints would be very much appreciated.

You have already proved that, for $x \in (-\pi,\pi)$, by using a geometric series result,
$$
T(x)= \frac{\pi}{\pi-x}
$$ then to obtain $T'(x)$ just use
$$
\left(\frac 1u\right)'=-\frac{u'}{u^2}.
$$
