Evaluating $\lim_{n\to\infty}\frac{2^{n^{1.001}}}{n!}$ I need to evalulate 

$$\lim_{n\to\infty}\frac{2^{n^{1.001}}}{n!}$$

I thought that the limit is $0$ as same as $\dfrac{2^{n}}{n!}$, but apparently it's $
\infty$ and I just can't find the way to prove it.
Please, help
 A: Hint:  Note that $n!\le n^n$, and take the logarithm of our expression.  This logarithm is $\ge (\ln 2)n^{1.001} -n\ln n$.
A: Here's my take: for $n$ big enough, $2^{n^\varepsilon}>n$ (this is a consequence of the fact that $(\log n)/n^{\varepsilon}\to0$ for any $\varepsilon>0$). Then
$$
\frac{2^{n^{1+\varepsilon}}}{n!}=\frac{2^{n\cdot n^{\varepsilon}}}{n!}
=\frac{(2^{n^\varepsilon})^n}{n!}=\frac{2^{n^\varepsilon}}n\frac{2^{n^\varepsilon}}{n-1}\cdots\frac{2^{n^\varepsilon}}2\frac{2^{n^\varepsilon}}1>2^{n^\varepsilon}\to\infty.
$$
Thus,
$$
\lim_{n\to\infty}\frac{2^{n^{1+\varepsilon}}}{n!}=\infty
$$
for all $\varepsilon>0$. 
A: Although properties and computations of logarithms (including exponents) are well developed in precalculus, more advanced techniques with logs and exponents such as derivatives, integrals, and yes limits too with natural logarithms do not come until much later in a first course in calculus with analytic geometry. Limits and simple derivatives are examined early in such a course which may explain Eyalbason's deficit in limits of logarithmic functions. So please try to exercise patience. We were all there at one time. 
