jmc

You can multiply the binomials in the order $(x-2)(x-2)(x+2)(x+2)$, but multiplying $(x-2)(x+2)$ first and then squaring the result means fewer terms to worry about since $-2x$ and $2x$ cancel each other. When we multiply $(x-2)(x+2)$, we get $x^2+2x-2x-4=x^2-4$. There is another set of $(x-2)(x+2)$ that also equals $(x^2-4)$. So, the simplified expression is $(x^2-4)(x^2-4)=x^4-8x^2+16$. The product of the coefficients is $1\cdot -8\cdot 16=\boxed{-128}$.