Prove or disprove $f(n)\in \omega(log_2(n)) \Rightarrow n\in o(2^{f(n)})$

Question

I can't prove it and I can't disprove it.
My try :
$f(n)>log_2(n)$ so $\frac {n}{2^{log_2(n)}} \ge \frac {n}{2^{f(n)}}$
$\lim \limits_{n \to \infty} \frac{n}{2^{log_2(n)}}=\lim \limits_{n \to \infty} 1=1$
which means $$0\le \lim \limits_{n \to \infty} \frac {n}{2^{f(n)}} \le1$$
But isn't necessary $0$ so there might be a counter-example.

Answer 1

The chaim is not true. Put $f(n) = -n$.

If you suppose that $f(n) > 0$ then the chaim is true. We have $\frac{f(n)}{\log_2 n} \to +\infty$, so for $n\ge N$ we have $\frac{f(n)}{\log_2 n} \ge 2$ and $$0 \le \frac{n}{2^{f(n)}} \le \frac{n}{2^{2\log_2 n}} = \frac{f(n)}{(2^{\log_2 n})^2} = \frac{n}{n^2} \to 0.$$