I need help with the following question:  If $1&lt;s\in \mathbb N$ and $p=2^s+1$ is prime, so $7$ is a primitive root modulo $p$.  My thoughts: First I know: $\phi(p)=p-1=2^s$. So if $r$ is the order of $7$ modulo $p$ then $r | \phi (p)=2^s$ then $rc=2^s$ and I don't know how to continue. Thank in advance.
What I want to show is the following statement.  For every prime of the form $2^{4n}+1$, 7 is a primitive root.  What I get is that  $$7^{2^{k}}\equiv1\pmod{p}$$ $$7^{2^{k-1}}\equiv-1\equiv2^{4n}\pmod{p}$$ $$7^{2^{k-2}}\equiv(2^{n})^2\pmod{p}$$ Thus $(\frac{2^n}{p})=(\frac{7^{2^{k-2}}}{p})=1$.  I think that $7$ is important because $7$ is a primitive root but I don't know how to use $7$.
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