For which $x\in\mathbb{R}$ is the series of general term $a_n = x^{n!}$ convergent? I firstly found the simplified form of $\frac{a_{n+1}}{a_n} = |x|\cdot|x^n|$ and used this to establish the end points $-1\lt x\lt 1$. I then tested the end points by finding the limit to infinity of $(-1)^{n!}$ and $1^{n!}$. By doing this I established that (by the divergence test) at neither end-point the series converged. Therefore I concluded that the interval of convergence is $(-1,1)$. I'm not sure if this is correct. If it's wrong can you please tell me where I went wrong (but don't tell me the full answer please!).

Your use of the ratio test is fine--clearly, it won't converge outside of $[-1,1]$, and you are correct in asserting that it doesn't converge for $x=-1$ and $x=1$.

You probably still have to prove that it converges for $x\in (-1,1)$, but you can use the comparison test against the series $a_n = x^n$ for that.