jmc

Applying difference of squares, we get $$\begin{array}{rl}& \frac{(1998^2-1996^2)(1998^2-1995^2)\cdots (1998^2-0^2)}{(1997^2-1996^2)(1997^2-1995^2)\cdots (1997^2-0^2)}\\ & =\frac{(1998+1996)(1998-1996)(1998+1995)(1998-1995)\cdots (1998+0)(1998-0)}{(1997+1996)(1997-1996)(1997+1995)(1997-1995)\cdots (1997-0)(1997+0)}\\ & =\frac{3994\cdot 2\cdot 3996\cdot 3\cdots 1998\cdot 1998}{3993\cdot 1\cdot 3992\cdot 2\cdots 1997\cdot 1997}.\end{array}$$In the numerator, we get every number from 2 to 3994 as a factor, with 1998 appearing twice. In the denominator, we get every number from 1 to 3993 as a factor, with 1997 appearing twice. Thus, the fraction simplifies to $$\frac{1998\cdot 3994}{1997}=\boxed{3996}.$$