I'm attempting to show that $2$ is not a primitive root of primes of the form $p = 8k + 7$. I know that, to do so, I must show that $2$ has order less than $\phi(p)$ modulo $p$ (where $\phi$ denotes Euler's Totient function), however I've found myself a bit stuck.
I've begun by way of contradiction. If we assume $2$ to be a primitive root of $p$, then:
$$2^{\phi(8k+7)} \equiv 1 \pmod{8k + 7}$$
Since $p$ is prime, $\phi(p) = (8k+7)-1 = 8k+6$, hence:
$$2^{8k+6} \equiv 1 \pmod{8k+7}$$
...it isn't clear to me where to go from here.