Time complexity – Big-O comparison of two algorithms $ sqrt {n} $ and $ 2 ^ { sqrt { log _ {2} n}} $


I have given 2 algorithms with their temporal complexity $ t_a (n) = sqrt {n} $ and $ t_b (n) = 2 ^ { sqrt { log _ {2} n}} $ and I have to show $ t_b (n) = O (t_a (n)) $,

I have created a program to check this statement and it seems that this is for everyone $ c> 0, forall n geq16 $ it applies, but I don't know how to prove it formally because I can't find a simplification for it $ t_b $,

I know I have to prove $ exists c
: forall n geq N: t_b (n) leq c * t_a (n) $
with big-o notation.

A hint / solution would be really great.