In number theory, Wilson’s theorem states that a natural number n > 1 is a prime number if and only if((n-1)!) mod n =(n-1)That is, it asserts that the factorial (n – 1)! = 1 * 2 * 3 * ……………*(n – 1) is one less than a multiple of n exactly when n is a prime number.for example :for n= 4(n-1) ! = 6(n-1) ! mod n =2for n= 5(n-1) ! = 24(n-1) ! mod n = 4for n=6(n-1) != 120(n-1) ! mod n =0for n=11(n-1) ! = 3628800(n-1) ! mod n = 10for n =12(n-1)! = 39916800(n-1) ! mod n = 0Proof:It is easy to check the result when n is 2 or 3, so let us assume n > 3. If n is composite, then its positive divisors are among the integers 1, 2, 3, 4, … , n-1 and it is clear that gcd( (n-1)! , n) > 1, so we can not have (n-1)! = -1 (mod n).However if n is prime, then each of the above integers are relatively prime to n. So for each of these integers a there is another b such that ab = 1 (mod n). It is important to note that this b is unique modulo n, and that since n is prime, a = b if and only if a is 1 or n-1. Now if we omit 1 and n-1, then the others can be grouped into pairs whose product is one showing2.3.4…..(n-2) = 1 (mod n)(or more simply (n-2)! = 1 (mod n)). Finally, multiply this equality by n-1 to complete the proof.Note:Wilson theorem holds only for prime numbers .Problem for practice :Factorial Again