# Extending an analytic function to an entire function

I encountered a problem from an old complex analysis prelim which I'm having a hard time with. The problem: Let $f(z)$ be entire, and $g(z)$ be an analytic function in a neighborhood of $z=1$ which satisfies $$g^{(n)}(1) = \frac{(f^{(n)}(1))^{\alpha}}{(n!)^{\alpha -1}}, \alpha >0.$$ Prove that $g$ can be extended to an entire function.

As an attempt, I tried doing this by recalling that an analytic function $g$ in a neighborhood of $z=1$ has a power series expansion centered at $1$ given by $$g(z)= \sum_{n=0}^{\infty} a_n (z-1)^n,$$ where $a_n= \frac{g^{(n)}(1)}{n!}$. I think the same should also be done for $f$? This is where I'm not sure how to continue, because if I recall an entire function is analytic on the whole complex plane so it also has a power series but with infinite radius of convergence, with center at $z=0$, or does it matter where the center of the series is? I don't think it should matter as I believe the power series expansion of an entire function can be centered at any $z \in \mathbb{C}$, correct?

If the answer to my above question is that an entire function can be centered at any point in the complex plane, then given by how the $n$th derivatives of $g$ at $1$ are defined, we would have that $a_n = (b_n)^{\alpha}$ where $b_n= \frac{f^{(n)}(1)}{n!}$. Am I right so far? Then, we can see that the radius of convergence of $g$ is the same as that of $f^{\alpha}$, i.e. $\infty$, since the radius of convergence of $f$ is infinite, given that $f$ was assumed entire. Now does this make any sense? Do I need to correct anything in my "understanding?" I would appreciate some input here.

Raising complex numbers to a potentially fractional power $\alpha$ is a bit tricky, but we can avoid this headache by focusing on $$|g^{(n)}(1)| = \frac{|f^{(n)}(1)|^{\alpha}}{(n!)^{\alpha -1}} \tag1$$ The key fact is that the property of being entire is equivalent to having infinite radius of convergence of power series. Which is itself equivalent to having $$\lim_{n\to\infty}\frac{|f^{(n)}(1)| }{n! } R^n =0 ,\qquad \forall R>0$$ or, using Landau notation, $|f^{(n)}(1)| /n! = o(R^{-n})$ for every $R>0$.
By virtue of (1), $$\frac{|g^{(n)}(1)|}{n!} = \frac{|f^{(n)}(1)|^{\alpha}}{(n!)^{\alpha ]}} = o(R^{-\alpha n})$$ Since this holds for every $R$, the radius of convergence is infinite.