Convergence radius of a power series I came across the following question :

Find the convergence radius for the series $\sum_{n=1}^\infty n!x^{n!}$.

My initial intuition led me to believe it should converge for $x\in(-1,1)$.

I would really appreciate if anyone could review my proof and point out if I`m missing anything.

The proof goes as following : 

Let $a_n$ be the sequence of coefficients of $x^n$. For every $n_m=k!$ for some $k\in \mathbb{N}$, $a_{n_m}=n_m$.

By Cauchy-Hadamard theorem, $\limsup (|a_n|^\frac 1 n)=\frac 1R$, and by D'Alembert criterion, if the limit $\frac {a_{n_m+1}} {a_{n_m}}$ exists , it equals to  $\lim (|a_{n_m}|^\frac 1 {n_m})$ , and indeed $\lim \frac {a_{n_m+1}} {a_{n_m}}=\frac {{n_m+1}} {{n_m}}=1$. Therefore we have $1$ as a partial limit of $|a_n|^{\frac 1 n}$. In addition, for every $n\neq n_m$, $a_n = 0$, meaning $1$ is the limit supremum. (The partial limits are either 1 or 0)

From here we can deduce R=1.

Divergence for $x=1$ is obvious, since $n!$ doesn't converge to $0$, and divergence for $x=-1$ for the same reason (since $n!$ are all even after $n=2$, we get the same series as $x=1$)

Is my proof correct?
Note that $$\sum_{n=1}^M n!x^{n!}\le \sum_{n=1}^{M!} n|x|^{n}$$

So what can we say about the limit case?
It certainly does NOT converge for $|x| \geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x \gt 0$) by $\sum mx^m.$