Proving a limit of a sequence I have to prove that for a > 1, $$\lim_{ n\rightarrow \infty  }{ { \left  (\frac { 2^{n^a} }{ n! } \right ) } } = \infty$$
I've tried to apply L'Hôpital's rule and d'Alembert's ratio test, but without any success...
Any help would be greatly appreciated!

Using that $(n-k)/n<1$ for all $k=1,\ldots,n-1$, we have (for $n$ big enough such that $n^{a-1}\log 2\geq \log n$)
$$
\frac{2^{n^a}}{n!}\geq\frac{n^n}{n!}=\frac1{1 \;\frac{n-1}n\;\cdots\;\frac1n}\geq\frac1{\frac1n}=n\to\infty
$$
