On more of those

As $\omega$ runs through all primitive $n$ th roots of unity (for odd $n$ ), so does $\omega^2$ . Thus $\prod(1-\omega)=\prod(1-\omega^2)$ $=\prod(1-\omega)\prod(1+\omega)$ so $\prod(1+\omega)=1$ . The square of the reciprocal is $P=\boxed{1}$

write as $\prod (-1-\omega)^{-2}$ a polynomial can be written as $p(x)=\prod(x-roots)$ . but what polynomial has roots $\omega?$ . the cyclotomic polynomial . we know $\Phi_n(x)=\prod_{d|n} \left(1-x^{n/d}\right)^{\mu(d)}$ putting 2015 =n and -1=x $\Phi_{2015}(-1)=\dfrac{(1-(-1)^{2015})(1-(-1)^5)(1-(-1)^{13})(1-(-1)^{31})}{(1-(-1))(1-(-1)^{65})(1-(-1)^{155})(1-(-1)^{403})}=\boxed{1}$ this raised to -2 is also 1.