It's easy to understand intuitively, but to provide a proof, I will be using a counterexample.if f(n) = 2(n^2 ) g(n) = (n^2)
2^f(n) = 2^(2(n^2)) and 2^g(n) = 2^(n^2)
According to big-Oh notation
f(n) <= cg(n) for all values of n >= no
but this isnt possible in this case as 2^(f(n)) = (4^(n^2)) grows faster than (2^g(n)) = (2^(n^2))