elementary number theory - What is the remainder of $18!$ divided by $437$?

What is the remainder of $18!$ divided by $437$?

I'm getting a little confused in the solution. It uses Wilson's theorem

Wilson's Theorem:
If $p$ is prime then $(p-1)!\equiv-1(\text{mod } p)$

So it first factors $437$ into primes. So $437 = 19 \cdot 23$. Then from Wilson's theorem notes that $18!\equiv-1(\text{mod } 19)$ so we're part way there, but also says $22\equiv22!(\text{mod }23)$ by Wilson's theorem (really don't know how they got this from $22!\equiv-1(\text{mod }23)$.

Also I'm confused how solving this leads to finding the remainder for $18!$ divided by $437$? I understand getting $18!$ from $19$ but not the $23$ part.

Answer

By Wilson's theorem, $18!\equiv-1\mod 19$ and $22!\equiv-1\mod 23$. Now

$22!=22\times21\times20\times19\times18!\equiv(-1)(-2)(-3)(-4)18!\equiv(24)18!\equiv(1)18!=18!\mod 23.$

Therefore $18!\equiv-1\mod19$ and $18!\equiv-1\mod 23$.

By the constant case of the Chinese remainder theorem, therefore,

$18! \equiv-1\equiv436\mod 437=19\times23$.