HINT $\rm\quad\displaystyle \frac{f(n)}{2^n}\ =\ \frac{2}2\ \frac{2}{2}\ \frac{3}2\ \frac{4}2\ \cdots\ \frac{n}2\ > 1\ $ for $\rm\ n > 2\ $ since each factor is $>1$ after the 2nd factor.
Generally that works to show that factorials grow faster than powers, i.e. $\rm\ f(n) > c^n\ $ for $\rm\ n > n_0\:.\ $ It suffices to show: eventually $\rm\ g(n) = f(n)/c^n > 1\:,\: $ or, equivalently, eventually $\rm\: g(n+1)/g(n) > 1 \ $ since, by multiplicative telescopy, $\rm\:g(n)\:$ is a product of these adjacent term ratios, namely
$\rm g(0)\ \ \prod_{k\:=\:0}^{n-1}\ \frac{g(k+1)}{g(k)}\ = \ \ {\rlap{--}g(0)}\frac{\rlap{--}g(1)}{\rlap{--}g(0)}\frac{\rlap{--}g(2)}{\rlap{--}g(1)}\ \ \cdots\ \ \frac{g(n)}{\rlap{----}g(n-1)}\ =\ \ g(n) $
Yours has $\rm\ g(k+1)/g(k)\ =\ (k+1)/2\ >\ 1\ $ for $\rm\ k > 1\ $ so $\rm\:g(n) > 1\:,\:$ as a product of terms $> 1\:.$
As I have emphasized here before in many posts, by means of cancelling complicated expressions, telescopy often reduces induction problems to trivialities (e.g. a product of terms $> 1$ is itself $> 1$). Difficult problems involving hyperrational functions (i.e. $\rm\ f(n+1)/f(n) = $ rational function of $\rm\:n\:,\:$ such as powers and exponentials) are, after application of telescopy, greatly simplified to trivial problems about rational functions - functions so simple that questions about such can be decided mechanically by algorithms.