Prove that the limit of $2^{\frac{-1}{\sqrt{n}}}=1$ 
Prove that the limit of $2^{\frac{-1}{\sqrt{n}}}=1$.

I need to show that for each $\epsilon$ there exists an $n_0 \in \mathbb{N}$ such that $ \forall n \geq n_0: |2^{\frac{-1}{\sqrt{n}}}-1|\lt \epsilon$
I was simply trying to solve $2^{\frac{-1}{\sqrt{n}}}=1$ by taking logs, by that doesn't lead my anywhere. Can anybody help please?

No, it isn't good way, because from $2^{\frac{-1}{\sqrt{n}}}=1$ you get $\frac{-1}{\sqrt{n}}=0$-it's not possible.
What you can do:
1) If you know that for all $a>0$ $\lim_{n \to \infty} a^{\frac{1}{n}}=1$ you can use this.
2)If you don't know that for all $a>0$ $\lim_{n \to \infty} a^{\frac{1}{n}}=1$ you can prove this (there is, for example, proof using Bernoulli's inequality. 
