Prove that, $\lim_{n \to \infty} n^{1 \over \sqrt n}= 1$

Question

Actually I have a problem of power series where I am asked to find the Radius of convergence of it. The power series looks like
$\sum_{n=1}^\infty {x^n \over n^\sqrt n} $
So I need to calculate the limsup of $n^{1 \over \sqrt n}$, as the sequence $\{n^{1 \over \sqrt n} \}$ converges I have just mentioned limit instead of limsup.
I have tried to prove the limit using a standard limit $\lim_{n\to\infty} n^{1 \over n}=1$ using squeeze test.
Can anybody assist me to prove the problem? Only hint will work for me.

Answer 1

Hint: $$n^{\frac{1}{\sqrt n}} = (\sqrt n)^{\frac{1}{\sqrt n}}\cdot (\sqrt n)^{\frac{1}{\sqrt n}}$$