Calculate 2000! (mod 2003) Calculate 2000! (mod 2003)
This can easily be solved by programming but is there a way to solve it, possibly with knowledge about finite fields? (2003 is a prime number, so mod(2003) is a finite field) .
As much details as possible please, I want to actually understand.

Wilson's theorem is your friend here.
$$(p-1)! \equiv -1 \mod p$$ for prime $p$.
Then notice $$-1 \equiv (2003-1)! = 2002 \cdot 2001 \cdot 2000! \equiv (-1) (-2) \cdot 2000! \mod 2003.$$
