How do I prove that if $p$ is prime then $p$ divides $2^{p}-2$? I know that if $p$ divides $2^{p}-2$ can be written as $2^p - 2 \equiv 0 \bmod p$, but then I get stuck. Im not sure how to take an approach on this. 

Fermat's little theorem states that $a^{p-1}-1$ is divisible by $p$ for any prime $p$ and any integer $a$.
