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?

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
$$
\frac{2^{n^2}}{n!}\ge n+1
$$
