Disprove $f(n) = 2^{2n}$ is $O(2^n)$
Informally I know that this means we have to show that $2^{2n}$ grows much faster than $2^n$.
More formally, I know that we need to show that for any $N$, there is a constant $c>0$, so that for all $n>N$ such that $2^{2n}>c\cdot 2^n$.
I am not sure how to derive $N$ and a fixed constant $c$ in order to disprove this proposition. Can anybody show me how?
Hint: $$\frac{2^{2n}}{2^n}=2^n$$