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?
 A: We have $$\frac{2^{n^2}}{n!}=\frac{(2^n)^n}{n!}>\frac{n^n}{n!}>n$$
A: Let's prove that $2^{n^2}\ge(n+1)!$, for every $n$. The case $n=0$ is obviously true. We also have 
$$
2^{(n+1)^2}=2^{n^2}\cdot 2^{2n+1}\ge 2^{2n+1}(n+1)!
$$
and we're done if we show that $2^{2n+1}\ge n+2$. Again, the base step is trivial; moreover
$$
2^{2n+3}=4\cdot 2^{2n+1}\ge 4(n+2)=n+3n+8\ge n+3
$$
Therefore
$$
\frac{2^{n^2}}{n!}\ge n+1
$$
A: 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
$$\frac{2^{n^2}}{n!} > \frac{2^{n^2}}{n^n} \ \ \implies \ \ \lim_{n \to \infty}\left[\frac{2^{n^2}}{n!}\right] = \infty$$
