Radius of convergence for complex series $\sum_{n=1}^\infty \frac{z^{n^2}}{n!}$ 
Find the radius of convergence for $$\sum_{n=1}^\infty \frac{z^{n^2}}{n!}$$ where $z \in \mathbb{C}$

It is immediately clear that $R \ge 1$, since for $z = 1$, the sum is just the Taylor expansion for $e$.
Viewing this sum as a power series of the form $\sum_{n=1}^\infty c_n z^n$, it is also clear that $c_{n^2} = \frac1{(n^2)!}$ and $c_n = 0$ if $n$ is not a perfect square. 
Applying the root test, the limit supremum is greater than or equal to $0$, so we need only check the case that $n$ is a perfect square:
From Sterling's Approximation $\frac1{n!}$ has the same limit behavior as $c(\frac{e}{n})^n\frac1{\sqrt{n}}$ for some constant $c$
So $\frac1{(n^2)!}$ has the same limit behavior as $c(\frac{e}{n^2})^{n^2} \frac1{n}$
This means that:
$$\lim_{n \to \infty} (|c_{n^2}|)^{1/n} = \bar{c} \lim_{n \to \infty} (\frac{e}{n^2})^n \frac1{n^{1/n}}$$
For some constant $\bar{c}$. But (to my understanding):
$$\lim_{n \to \infty} \frac1{n^{1/n}} = 1$$ 
and
$$ \lim_{n \to \infty} (\frac{e}{n^2})^n = 0$$
so we get that $\lim\sup (|c_n|)^{1/n} = 0$, and so $R = \infty$.
However, I have serious doubts of the validity of my answer. I suspect that for $|z|>1$, the sum diverges as I believe the limit of the argument tends towards infinity.
Can anybody point out my mistake for me?
Edit:
As pointed out in the comments,
$c_{n^2} = \frac1{n!}$, adjusting this and the power of the root: 
$$\lim_{n \to \infty} (|c_{n^2}|)^{1/n^2} = \lim_{n \to \infty} (\frac1{n!})^{1/n^2} = \lim_{n \to \infty} c^{1/n^2} (\frac{e}{n})^{1/n} \frac1{n^{1/2n}} = 1$$

Your answer is incorrect. One can use limit comparison to show that this series diverges for $|z|>1$, without even considering it like a power series:
$$\frac{z^{n^2}}{n!} \sim \frac{z^{n^2}}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}.$$
$$\frac{z^{n^2}}{n!} \sim \frac{1}{\sqrt{2\pi n}}\left(e\frac{z^n}{n}\right)^n.$$
If $z>1$, $\frac{z^n}{n}$ is unbounded, which means the terms themselves are unbounded (this is a bit hand-wavy, but it's correct).
