What is $O(\sqrt{2^n}n^2)$?

Question

What is $O(\sqrt{2^n}n^2)$?

Is it $O(2^n)$, or does the square root cause it to be reduced? I'm trying to analyze an algorithm that I came up with, and if it still has exponential time cost, I'm going to have to try harder.

Thanks!

Answer 1

$\sqrt {2^n}=2^{\frac n2}$, still exponential in the sense that it will dominate $n^k$ for any $k$