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!

Ratio test works fine here. We have $$x_n = \dfrac{2^{n^a}}{n!} \implies \dfrac{x_{n+1}}{x_n} = \dfrac{2^{(n+1)^a}}{2^{n^a}} \cdot \dfrac{n!}{(n+1)!} = \dfrac{2^{(n+1)^a - n^a}}{n+1}$$
Now from binomial theorem, we have that
$$(n+1)^a \geq n^a + an^{a-1} \implies (n+1)^{a} - n^a \geq an^{a-1}$$
Hence,
$$\dfrac{x_{n+1}}{x_n} \geq \dfrac{2^{an^{a-1}}}{n+1} \to \infty \text{ since }a>1$$
