On the growth order of an entire function $\sum \frac{z^n}{(n!)^a}$ Here $a$ is a real positive number. The result is that $f(z)=\sum_{n=1}^{+\infty} \frac{z^n}{(n!)^a}$ has a growth order $1/a$ (i.e. $\exists A,B\in \mathbb{R}$ such that $|f(z)|\leq A\exp{(B|z|^{1/a})},\forall z\in \mathbb{C}$). It is Problem 3* from Chapter 5 of E.M. Stein's book, Complex Analysis, page 157. Yet I don't know how to get this. Will someone give me some hints on it? Thank you very much.

There is a formula expressing the growth order of entire function $f(z)=\sum_{n=0}^\infty c_nz^n\ $ in terms of its Taylor coeffitients:
$$
\rho=\limsup_{n\to\infty}\frac{\log n}{\log\frac{1}{\sqrt[n]{|c_n|}}}.
$$  
