Problem set 4 only self-inverses. Consequently, we can pair each of the remaining terms with its multiplicative inverse. Since the product of a number and its inverse is congruent to 1, all of these remaining terms cancel. Therefore, we have (mod p (mod P)4 Problem Set 4 only self­inverses. Consequently, we can pair each of the remaining terms with its multiplicative inverse. Since the product of a number and its inverse is congruent to 1, all of these remaining terms cancel. Therefore, we have: (p − 1)! ( ≡ 1 · p − 1) (mod p) ≡ −1 (mod p)