The smallest prime divisor of 18! + 1 So I have the following question:
Find the smallest prime divisor of $18!+1$.
In order to solve this I have used Wilson's Theorem as follows:
$$18!=(19-1)!$$
So we have $$(19-1)!+1$$
By Wilson's Theorem $${(19-1)!}\equiv{-1}\pmod{19}$$
And hence we have $$-1+1=0\pmod{19}$$
So $19$ is a divisor of $18!+1$.
Via a prime factorisation calculator I have found that this is in fact, the smallest prime divisor however I'm unsure of how to show this?

$$S=18!+1=2.3.4...18+1$$
if $a\in\{2,3,5,7...17\}$ is a prime which divides $ S, $ then it will divides  $ 1$ because it divides $18!$.
So $19$ is the smallest.
