TITLE: Prove that $2AB$ is square
QUESTION [6 upvotes]: Let $$A= 1! \cdot 2! \cdot 3! \cdots 1002!$$ 
$$B= 1004!\cdot 1005! \cdots 2006!$$
Prove that $2AB$ is square. Help guys, I tried, I really did but I couldn't.

REPLY [2 votes]: With help from above comment we have: 
\begin{align}
2AB &= 2(1!×2!×...×1002!)(1004!×1005!×...×2006!) \\
&= 2(2^{501}×1×2×...×501)(2^{500}×502×503×...×1003)((1!)^2(2!)^2 ...(1003!)×(1005!)^2 ...(2005!)^2 \\
&= 2^{1002}(1!)^2 ...(1003!)(1003!)(1005!)^2 (1006)^2 ...(2005)^2 \\
&=  2^{1002} ...
\end{align}
square.