2015/091) Let p be a prime number. Prove that 6[(p-4)!] = 1 (mod p )

we have 6 = 2 * 3
p is prime
now (p-2) mod p = -2
(p-3) mod p = - 3

so 6(p-4)! = - (p-2)(p-3) (p-1)(p-4)! = - (p-1)!

so 6(p-4)! mod p = - 1(p-1)! mod p = -1 * (-1) = 1 mod p