For every prime of the form $2^{4n}+1$, 7 is a primitive root. 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 $7$.
 A: Assume that $p=2^{4n}+1$ is a prime. We know that the order of $7$ is a factor of $p-1$, so it is a power of two. The claim is equivalent to saying that the order is exactly $p-1$. Assume that this is not the case. Then the order is a factor of $(p-1)/2$ meaning that
$$
\left(\frac7p\right)\equiv 7^{(p-1)/2}\equiv 1\pmod p.
$$
The case $n=0$ is easy, so we can assume that $n>0$. Then $p\equiv1\pmod4$, so the law of quadratic recpirocity tells that
the claim is equivalent to
$$
\left(\frac{p}7\right)=1.
$$
As $2^3\equiv1\pmod7$, $p\equiv 2^n+1\pmod7.$ The residue class of $2^n+1$ modulo $7$ can be either $2,3$ or $5$, when $n\equiv 0,1,2\pmod3$ respectively. Of these, only $2$ is a quadratic residue modulo $7$. This means that we must have $3\mid n$. I leave it to you to prove that in that case $p$ cannot be a prime unless $n=0$ and $p=2$.
A: Case $n=0$ is trivial and satisfies our assumption.
So consider all possible outcomes when the natural number $n = 3k + t$ for some $k \in \mathbb{Z}$ and $t \in \{\, 0,1,2 \,\}$:
\begin{align*}
& p = 2^{4\cdot 3k} + 1 = 2^{12k}+1 = (2^{4k} + 1)(2^{8k} - 2^{4k} + 1) \not\in \mathbb{P} \quad (\text{prime}); \\
& p = 2^{4\cdot (3k+1)} + 1 = 8^{4k} \cdot 16^1  + 1 \equiv 1^{4k} \cdot 2^1 + 1 = 3 \mod{7}; \\
& p = 2^{4\cdot (3k+2)} + 1 = 8^{4k} \cdot 16^2  + 1 \equiv 1^{4k} \cdot 2^2 + 1 = 5 \mod{7}.
\end{align*}
If $p \equiv 3 \mod{7}$ then we have $(p/7) = (3/7) = -(7/3) = -(1/3) = -1$.
If $p \equiv 5 \mod{7}$ then we have $(p/7) = (5/7) = (7/5) = (2/5) = -1$.
Since obviously $p \equiv 1 \mod{4}$, quadratic reciprocity law gives $(7/p) = (p/7)$ and
\begin{equation*}
-1 = (p/7) = (7/p) \equiv 7^{\frac{p-1}{2}} = 7^{2^{4n-1}} 
\mod{p}.
\end{equation*}
This means that $\operatorname{ord}_p(7) \nmid 2^{4n-1}$ while at the same time $\operatorname{ord}_p(7) \mid \varphi(p) = 2^{4n}$.
Then the only possibility is that $\operatorname{ord}_p(7) = \varphi(p)$ and $7$ is a primitive root modulo $p$ by definition.
A: Recall that any  prime $p$ has $\varphi(p-1)$ primitive roots. In our case, $\varphi(p-1)=(p-1)/2$. 
Any prime $p$ has $(p-1)/2$ quadratic non-residues. Any primitive root is a non-residue, so for primes $p$ of the form $2^w+1$, every NR is a primitive root. 
It remains to show that $7$ is a NR of $p$. This is dealt with in the answer by Jyrki Lahtonen. Briefly, use Reciprocity. 
