jmc

We can apply difference of squares to the numerator: $$n^2+2n-1=(n+1{)}^2-2=(n+1+\sqrt{2})(n+1-\sqrt{2}).$$We can also factor the denominator: $$n^2+n+\sqrt{2}-2=(n+\sqrt{2})+(n^2-2)=(n+\sqrt{2})+(n+\sqrt{2})(n-\sqrt{2})=(n+\sqrt{2})(n-\sqrt{2}+1).$$Hence, $$\frac{n^2+2n-1}{n^2+n+\sqrt{2}-2}=\frac{(n+1+\sqrt{2})(n+1-\sqrt{2})}{(n+\sqrt{2})(n-\sqrt{2}+1)}=\frac{n+1+\sqrt{2}}{n+\sqrt{2}}.$$Therefore, $$\begin{array}{rl}\prod _{n=1}^{2004}\frac{n^2+2n-1}{n^2+n+\sqrt{2}-2}& =\prod _{n=1}^{2004}\frac{n+1+\sqrt{2}}{n+\sqrt{2}}\\ & =\frac{2+\sqrt{2}}{1+\sqrt{2}}\cdot \frac{3+\sqrt{2}}{2+\sqrt{2}}\cdot \frac{4+\sqrt{2}}{3+\sqrt{2}}\cdots \frac{2005+\sqrt{2}}{2004+\sqrt{2}}\\ & =\frac{2005+\sqrt{2}}{1+\sqrt{2}}\\ & =\frac{(2005+\sqrt{2})(\sqrt{2}-1)}{(1+\sqrt{2})(\sqrt{2}-1)}\\ & =\frac{2004\sqrt{2}-2003}{1}\\ & =\boxed{2004\sqrt{2}-2003}.\end{array}$$