modulo group of prime order $p$ is prime of the form $p=2^{n}+1$ for $n\in N$. How can I show that $o(2)=2n$ for $2\in Z_{p}$?
And how can I proove that $n=2^{k}$ for some $k\in N$?
For the first, the only clue that I see that from theorem: $$$$
$(2,p)=1$ then $2^{\phi(p)}=1$ and $\phi(p)=p-1$ then $\phi(2^{n}+1)=2^n$
so generally: $$2^{2^{n}}=1$$
Edited: Second part of the question
$$n=2^{k}|k\in N$$
How do I prove that ?

Hint: $2^n\equiv -1$ mod $(2^n+1)$.
