Verified answer

Recall Wilson's theorem, which states that for a prime p, (p - 1)! = -1 (mod p).

Additionally, note that we need q > 2.

So, for q prime,

2(q - 3)! + 1 = n (mod q)

2(q - 3)! = n - 1 (mod q)

2(q - 3)!(q - 2)(q - 1) = (n - 1)(q - 2)(q - 1) (mod q)

2(q - 1)! = (n - 1)(-2)(-1) (mod q)

-2 = 2(n - 1) (mod q)

As gcd(2,q) = 1, 2 has a multiplicative inverse modulo q. Let this inverse be m.

-2m = 2m(n - 1) (mod q)

-1 = n - 1 (mod q)

0 = n (mod q)

Hence, 2(q - 3)! + 1 = 0 (mod q).

You cannot use induction with primes, only with ordinary naturals or integers.

Try it by using the Fermat's Little Theorem.

For a q prime and b a positive integer not divisible by q, a^(q-1)-1 ≡ 0 (mod q)