Prove that interval of convergence for Σ(n!x^(n!)), n=1..∞ is (-1,1)

I need to prove that the interval of convergence for the infinite sum of n!x^(n!) going from n to infinity is (-1,1). I tried the ratio test which doesn't seem to work in this case. Any help would be greatly appreciated!

Using the ratio test:

r = lim(n→∞) |(n+1)! x^((n+1)!) / [n! x^(n!)]|
= lim(n→∞) (n+1) |x|^((n+1)! - n!)
= lim(n→∞) (n+1) |x|^(n! [(n+1) - 1)])
= lim(n→∞) (n+1) |x|^(n * n!).
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If |x| < 1, then holding x fixed:
Note that 0 ≤ |x|^(n * n!) < |x|^n.
==> 0 ≤ (n+1) |x|^(n * n!) < (n+1) |x|^n

Since lim(n→∞) (n+1) |x|^n
lim(n→∞) (n+1) / (1/|x|)^n; now of the form ∞/∞
= lim(n→∞) 1 / (1/|x|)^n * ln(1/|x|)], via L'Hopital's Rule
= lim(n→∞) |x|^n / ln(1/|x|)
= 0, since |x| < 1.

Hence, r = lim(n→∞) (n+1) |x|^(n * n!) = 0, by the Squeeze Theorem.
Since r < 1, the series converges.
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If |x| > 1, then r = ∞. So, the series diverges.
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At the endpoints x = -1 and 1, note that lim(n→∞) n! = ∞.
Hence, these series diverge by the nth term test.

So, the interval of convergence is (-1, 1).

I hope this helps!