i need to calculate the limit of this sequence:

$\displaystyle \sqrt[n^2]{n!}$

i tried to use the sandwich rule to squeeze it between to sequences, but got stuck.

this is what i did so far:

$\displaystyle \sqrt[n^2]{n!}<\sqrt[n^2]{n\cdot n\cdot \cdot \cdot \cdot \cdot n}=\sqrt[n^2]{n^n}=\sqrt[n]{n}\rightarrow 1$

but i couldn't find anything smaller then the given sequence that can be easily calculated.

for example, i know that:

$\displaystyle \sqrt[n^2]{n}<\sqrt[n^2]{n!}$

and i know that $\displaystyle \sqrt[n]{n}\rightarrow 1$, but is it enough to say that if $\displaystyle \sqrt[n]{n}\rightarrow 1$ then also $\displaystyle \sqrt[n^2]{n}\rightarrow 1$?

thanks in advanced!!!

BTW, if there's any other way to get to the answer ($\displaystyle \sqrt[n^2]{n!}\rightarrow 1$) i'd love to hear.