$2^{\sqrt{n}}>n^2$ inequality proof I am trying to do some trick with lim but get stuck on $2^{\sqrt{n}}>n^2$. I want to prove it for n large enough but don't know how to do induction step. Maybe this fact does not require a proof at all...

Taking for example $n=400$ as a base case
$$2^\sqrt{400}=2^{20}>10^6>400^2=16\cdot 10^4,x=\sqrt{n}\\2^x>x^4\\2^{x+1}>(x+1)^4\\2\cdot 2^x>x^4+4x^3+6x^2+4x+1\\x^4+15x^3>x^4+4x^3+6x^2+4x+1\\2^x+15x^3>x^4+15x^3\\\frac{3}{4}x^4\geq15 x^3\\2\cdot2^x>2^x+\frac{3}{4}2^x>x^4+4x^3+6x^2+4x+1$$
The $\frac{3}{4}x^4\geq15 x^3$ comes since $x\geq 20$
