1. Another congruence proof

Let p be a prime such that p >= 5. Show that 6 * (p-4)! -1 = 0 (mod p)

Thanks

2. By Wilson's Theorem $(p-1)!\equiv-1 \bmod{p}$

Now notice that $p-1\equiv-1,\; p-2\equiv-2,\; p-3\equiv-3 \bmod{p}$.

So $-1\equiv(p-1)! = (p-1)(p-2)(p-3)(p-4)! \equiv (-1)(-2)(-3)(p-4)! \bmod{p}$

Can you take it from here?

3. Yes!

Thank you so much. That was genius.