Problem: Let $$f(z) := \sum_{n=1}^\infty \frac{1}{n^4}z^{2^n}$$ noting that the sum has radius of convergence 1. Prove that there exists no neighbourhood U of 1 and holomorphic function g:$U\rightarrow\mathbb{C}$ such that f = g on $U\cap\mathbb{D}$.
Attempts: I have only vague ideas so far. My original idea was to use the uniqueness of power series and the fact that $g$ would be analytic at 1 to derive a contradiction of the fact that the radius of convergence of $f$ is 1. Help is appreciated!
$f$ is holomorphic on $\Bbb D$, so we have the expansion for $f'$, $$f'(x) = \sum_{n=1}^∞ \frac{2^n}{n^4} x^{2^n-1}$$ $f'$ has a non-removable singularity at $x=1$ because $$\lim_{\substack{x→ 1^- \\[0.1em] x∈ \Bbb R}}f'(x) = \sum_{n=1}^∞ \frac{2^n}{n^4} = ∞$$ so it cannot be extended to a holomorphic function on a neighbourhood including $1$, which means neither can $f$.