By induction, .
Then , so is convergent.
sqrt(2), sqrt(2*sqrt(2)), sqrt(2*sqrt(2*sqrt(2))), ...
converges, and find the limit.
So, uh, to show that it converges, we can use the monotone convergence theorem, that says if its bounded and monotone, it will converge... maybe?
Oh, maybe use the theorem that says that subsequences of a convergent sequence converge to the same lim as the orig. sequence? I'm not too sure.
We have the sequence which is strictly increasing and bounded (this would require a rigorous proof...but since you just said show, I'll show it in a simple way).
See the pattern?
Use the log rule:
We have: . Divide both sides by and thus: .
Take the limit as n goes to infinity: