jmc

We can write $r^5-1=0,$ which factors as $$(r-1)(r^4+r^3+r^2+r+1)=0.$$Since $r\ne 1,$ $r^4+r^3+r^2+r+1=0.$

To compute the product, we can arrange the factors in pairs: $$\begin{array}{rl}(r-1)(r^2-1)(r^3-1)(r^4-1)& =[(r-1)(r^4-1)][(r^2-1)(r^3-1)]\\ & =(r^5-r-r^4+1)(r^5-r^2-r^3+1)\\ & =(1-r-r^4+1)(1-r^2-r^3+1)\\ & =(2-r-r^4)(2-r^2-r^3)\\ & =4-2r^2-2r^3-2r+r^3+r^4-2r^4+r^6+r^7\\ & =4-2r^2-2r^3-2r+r^3+r^4-2r^4+r+r^2\\ & =4-r-r^2-r^3-r^4\\ & =5-(1+r+r^2+r^3+r^4)=\boxed{5}.\end{array}$$