I'm trying to show that $|f(r)| \rightarrow \infty$ as $r \rightarrow 1^-$. Here, $f(z) = \sum_{n\geq 0}z^{2^n}$.
Then I need to show by induction that if $e^{i\theta 2^m} = 1$ for a certain $m$ in $\mathbb{N}$ then $|f(re^{i\theta})| \rightarrow \infty$ if $r\rightarrow 1^-$.
For this one, I imagine the case $m = 0$ would be hard, but by induction, if we assume this is true for $m$, then showing for $m + 1$ is relatively simple since it boils back down to the first question since:
$re^{i2^{m+1}\theta} = re^{i2^m\theta * 2} = r(e^{i2^m\theta})^2 = r * 1^2 = r$
Or am I wrong?