Does $\lim_{n \rightarrow \infty} \frac{2^{n^2}}{n!}$ exist?

We have
\[\frac{2^{n^2}}{n!}\geq \frac{2^{n^2}}{n^n}=(\frac{2^n}{n})^n.\]
Since $\lim_{n\rightarrow \infty}\frac{2^n}{n}=\infty$, and hence
\[\lim_{n\rightarrow \infty}(\frac{2^n}{n})^n=\infty.\]
By comparision test, it follows from the above inequality that 

\[\lim_{n\rightarrow \infty} \frac{2^{n^2}}{n!}=\infty. \]