The smallest prime divisor of 18! + 1

Question

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?

Answer 1

$$S=18!+1=2.3.4...18+1$$ 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.