Let $\sum_{n=0}^{\infty}a_{n}T^{n} \in \mathbb{C}[[T]]$ be a nonzero power series over $\mathbb{C}$ with positive radius of convergence $R \in (0,\infty]$, and $f:B(0,R) \to \mathbb{C}$ be the analytic function $f(z):= \sum_{n=0}^{\infty}a_{n}z^{n}$. Suppose $f(0) \neq 0$, and that for any $z \in B(0,R/2)$, we have $f(2z)=f(z)^2$ in $\mathbb{C}$. How do we go about showing this: for any $z \in B(0,R)$,
$$f(z) = e^{bz}, \text{ where } b:=f'(0)=a_{1}?$$
I tried proving this directly by looking at the Taylor expansion of $\log$, as well as looking at the derivative of $f$, neither of which seem to work...
First $f(0)=f^2(0)$ and since $f(0) \ne 0, f(0)=1$.
Since $f(z)=0$ implies $f(\frac{z}{2})=0$ so any putative zero in $B(0,R)$ gives rise to an accumulating (at $0$) sequence of zeroes hence in particular to $f(0)=0$ which is not allowed, we get that $f \ne 0$ in $B(0,R)$ so an analytic $g=\log f$ exists and we fix it by choosing $g(0)=\log 1=0$ the principal one at $0$.
But then $e^g=f$ means $e^{g(2z)}=e^{2g(z)}$ so $g(2z)-2g(z)=2k(z)\pi i, k(z) \in \mathbb Z, z,2z \in B(0,R)$ and since $k(z)$ is then continuos it is constant $k(0)=0$ by our choice of $g(0)$.
Differentiating, we get $g'(2z)=g'(z)$ whenver $z,2z \in B(0,R)$ hence $g'$ constant since it has an accumulating sequence (at $0$) of equal values, so $g(z)=bz$, $f(z)=e^{bz}$ and clearly $b=f'(0)$ so we are done!