Prove that, $(2\cdot 4 \cdot 6 \cdot ... \cdot 4000)-(1\cdot 3 \cdot 5 \cdot ...\cdot 3999)$ is a multiple of $2001$ Prove that the difference between the product of the first 2000 even numbers and the first $2000$ odd numbers is a multiple of $2001$. Please show the method.
I have started with the following process:
$$(2\cdot 4 \cdot 6 \cdots 4000)-(1\cdot 3 \cdot 5 \cdots 3999)$$ 
How we can proceed it to find that it is a multiple of $2001$?

Hint: show that $2001$ is a factor of both products. One is very easy. For the other you need to know that $2001$ is not prime (you should be able to spot an easy factor).
