How to show that $\lim_{n \to \infty}\frac{2^{n^2}}{n!} = \infty$? I know $\lim \limits _{n \to \infty}\frac{2^{n^2}}{n!} = \infty$, but I need to prove it using the definition of limit, show that there is a $M$ such that all $a_n>M$.
I tried looking at $\frac{a_{n+1}}{a_n}$ and found out that $a_{n+1}>\frac{1}{2}a_n$, but I don't know then how to proceed, can someone help?

First notice that $n! < n^n$ and so
$$\frac{2^{n^2}}{n!} > \frac{2^{n^2}}{n^n}$$
Next, notice that $2^{n^2} \equiv (2^n)^n$, and so
$$\frac{2^{n^2}}{n^n} = \left(\frac{2^n}{n}\right)^{\! n}$$
Notice that $2^n \equiv (1+1)^n$, and taking the binomial expansion gives
$$2^n = 1 + n + \frac{n(n-1)}{2!} + \cdots$$
It follows that
$$\frac{2^n}{n} = \frac{1}{n} + 1 + \frac{n-1}{2!} + \cdots$$
We can then conclude that $\frac{2^n}{n} \to \infty$ as $n \to \infty$, as so
$$\lim_{n \to \infty}\left[\frac{2^{n^2}}{n^n}\right] = \lim_{n \to \infty}\left[\left(\frac{2^n}{n}\right)^{\! n}\right] = \infty$$
Furthermore
$$\frac{2^{n^2}}{n!} > \frac{2^{n^2}}{n^n} \ \ \implies \ \ \lim_{n \to \infty}\left[\frac{2^{n^2}}{n!}\right] = \infty$$
