For all primes p, show that $\displaystyle x^8 \equiv 16 \mbox{ (mod p)}$ is solvable.

From a theorem, the question is equivalent to showing that

$\displaystyle \displaystyle{16^{\frac{p-1}{(8, p-1)}} \equiv 1 \mbox{ (mod p)}}$

And then I am stuck.