If P congruent to $-1 \bmod$ 8 then $2^{(p-1)/2} - 1$ divisible by $p$ I am trying to prove the following statement:

If $p$ congruent to $-1 \bmod 8$, then $$2^{\frac{p - 1}{2}} - 1$$ divisible by $p$.

I can assume $p$ is congruent to $-1 \bmod 8$, then I must prove 
$2^{\frac{p - 1}{2}} - 1$ is divisible by $p$.
So $p + 1 = 8n$, but I'm confused how I can use this assumption to prove  $2^{\frac{p - 1}{2}} - 1$ divisible by $p$.

Let $p=8k-1$ and let $n=(p-1)/2 = 4k-1$.  Then note that
$$1\cdot 2\cdot 3\cdots n \equiv (-(p-1))\cdot 2 (-(p-3)) \cdot 4 (-(p-5)) \cdots (n-1)(-(p-n)) \pmod{p},$$
where we replaced each odd number $x$ by a congruent even number $-(p-x)$.
Now count the minus signs in the last product. Because $n$ is odd, there are $(n+1)/2 = 2k$ of them, so their product equals $1$.  So the congruence above becomes
$$n! \equiv (p-1)\cdot 2 (p-3) \cdot 4 (p-5) \cdots (n-1)(p-n) \pmod{p}.$$
Next, note that $2n = p-1$, $2(n-1) = p-3$, $2(n-2) = p-5, \ldots, n+1=4k =p-n,$ so we have
$$n! \equiv (2n)\cdot 2\cdot(2(n-1))\cdot 4 \cdots (n-1)\left(2\frac{n+1}{2}\right) \pmod{p}.  $$
Rearrange the right side to get
$$n! \equiv 2\cdot 4 \cdot 6 \cdots (n-1)(n+1)\cdots (2n) \equiv 2^n n!  \pmod{p}.$$
Cancel the $n!$ and you're done.
