Particularly, I want to prove $18!+1\equiv 0\pmod {23}$ so I wish to know if determining inverse of some numbers lead me to the proof correctly.
Give problem asks to Prove that $18!+1\equiv 0\pmod {437}$
$437$ is not a prime number but its factors $19$ and $23$ are prime.
Therefore using (in case of prime numbers only) Wilson's Theorem it can be shown that $18!+1\equiv 0\pmod {19}$.
To Prove $18!+1\equiv 0\pmod {23}$, we follow Wilson's theorem
$$(23-1)!\equiv -1\pmod{23}\\\Longrightarrow 21!\equiv 1\pmod{23},\space \gcd(22,23)=1 \to (I)$$.
I have determined inverses of $21,20,19$.
$$21x\equiv1\pmod{23} \to (i)\\20x\equiv1\pmod{23} \to (ii)\\19x\equiv1\pmod{23} \to (iii)$$ From $(i)$ $$\gcd (21,23)=1,\\23=21.1+2\\21=2.10+1\\\therefore 1=21-2.10\\=21-(23-21.1).10\\=21.11+23.(-10)$$
So from $(i), (ii), (iii)$ we have $11$ is the inverse of $21$, $(-8)$ is the inverse of $20$ and $(-6)$ is the inverse of $19$ and applying that in $(I)$ we get $$21!\equiv 1\pmod{23}\\\Longrightarrow 21.11.(20!)\equiv 11\pmod{23}\\\Longrightarrow 20!\equiv 11\pmod{23}\\\Longrightarrow 20.(-8).(19!)\equiv 11.(-8)\pmod{23}\\\Longrightarrow 19!\equiv 11.(-8)\pmod{23}\\\Longrightarrow 19.(-6).(18!)\equiv 11.(-8).(-6)\pmod{23}\\\Longrightarrow 18!\equiv 11.(-8).(-6)\pmod{23}\\\Longrightarrow 18!\equiv 11.48\pmod{23}\\\Longrightarrow 18!\equiv 11.2\pmod{23}\\\Longrightarrow 18!\equiv 22\pmod{23}\\\Longrightarrow 18!\equiv -1\pmod{23}\\\Longrightarrow 18!+1\equiv 0\pmod{23}$$
If this approach is correct then I can conclude that $18!+1\equiv 0\pmod{437}$. Any help is precious for learning and greatly appreciated. Any alternative quick method to solve this problem is also valuable.
