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.

For any odd prime $p$ we have $\left(p-1\right)!\equiv p-1\,\left(\text{mod}\,p\right)$ and $\left(p-2\right)\left(p-3\right)\equiv 2\,\left(\text{mod}\,p\right)$ so $\left(p-1\right)!\equiv \frac{p-1}{2}\,\left(\text{mod}\,p\right)$. The case $p=2003$ gives $\frac{p-1}{2}=1001$.
