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.

@Srivatsan's answer only proves the order of $f$ is at most $1 / \alpha$, so here's a way of proving that the order of $f$ is at least $1 / \alpha$. Consider the sequence $z_k := (ek)^\alpha$, and observe that
$$
|f(z_k)| = |f((ek)^\alpha)| = \sum_{n = 0}^\infty \frac{((ek)^\alpha)^n}{(n!)^\alpha}
\geq \frac{((ek)^\alpha)^k}{(k!)^\alpha} = \left(\frac{(ek)^k}{k!}\right)^\alpha.
$$
Using the inequality that $k! \leq k^k$, we get
$$
|f(z_k)|
\geq \left(\frac{(ek)^k}{k^k}\right)^\alpha
= e^{\alpha k}
= \exp\left(\frac{\alpha}{e} (z_k)^{1 / \alpha}\right).
$$
It then follows that $f$ cannot have order less than $1 / \alpha$.
