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.

$\mathbb{Z}_{2003}$ is a finite field. The equation $x^2 = 1$ has exactly two roots in that field: $x_1 = 1$ and $x_2 = -1 = 2002$. Thus, every $n \in \mathbb{Z}_{2003}^* \setminus \{ 1, 2002 \}$ has $n^{-1} \neq n$. Hence
$$\prod_{n=2}^{2001} n = 1,$$
because we can split the product into $1000$ pairs of form $(n, n^{-1})$ and the product of each pair cancels out. Therefore $2000! = (2001)^{-1} \pmod{2003}.$ 
