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. 

$$(2\cdot1)\cdot(2\cdot 2)...(2\cdot (p-1))=1\cdot2 ...\cdot (p-1)\ \ \ \ (\text{ mod } p )$$
Where the factors on the left and right are equal but not necessarily in the order written.

The reason is that the remainders of $2\cdot i$ mod $p$ are all different (and different to zero) for different $1\leq i\leq p-1$ and therefore are all the numbers $1,2,...,p-1$.

Then
$$2^{p-1}\cdot(p-1)!=(p-1)!\ \ \ \ (\text{ mod }p )$$
$$2^{p-1}=1\ \ \ \ (\text{ mod }p )$$
$$2^{p}=2\ \ \ \ (\text{ mod }p )$$
