jmc

Simplifying the squares, we have $$\begin{array}{rl}& (2^{1004}+5^{1005}{)}^2-(2^{1004}-5^{1005}{)}^2\\ & =2^{2008}+2\cdot 2^{1004}\cdot 5^{1005}+5^{2010}\\ & -2^{2008}+2\cdot 2^{1004}\cdot 5^{1005}-5^{2010}\\ & =4\cdot 2^{1004}\cdot 5^{1005}\end{array}$$Since $4\cdot 2^{1004}=2\cdot 2^{1005}$, we can rewrite the expression as $$2\cdot 2^{1005}\cdot 5^{1005}=2\cdot 10^{1005}=20\cdot 10^{1004}$$Thus, $k=\boxed{20}$.