Prove $$\log_2\left(\prod_{a=1}^{2015}\prod_{b=1}^{2015}\left(1+e^{\frac{2\pi i ab}{2015}}\right)\right)=13725,$$ where $i=\sqrt{-1}$.
My questions:
1: What does double product sign denotes? Does it denote that for each value of a the product for runs for $2015$ times for $b$?
2: I attempted by converting the terms to cosine product but could proceed no further.
Denote $\xi=e^{\frac{2\pi i~~}{2015~~~~}}$.
Proof:
Let $q=\frac{2015}m$, $$f(z):=z^{q}-1=\prod_{k=1}^{q}\left(z-\xi^{nk}\right).$$ The last equality is true because of the periodicty of primitive root of unity. For example,
Assume $\omega=e^{\frac{2\pi i~~~}{3~}}$, notice $({\color{red}2},3)=1$, we have $$z^3-1=(z-\omega^1)(z-\omega^2)(z-\omega^3)=(z-\underbrace{\omega^{{\color{red}2}\cdot 1}}_{=\omega^2})(z-\underbrace{\omega^{{\color{red}2}\cdot2}}_{=\omega^1})(z-\underbrace{\omega^{{\color{red}2}\cdot3}}_{=\omega^3}).$$
Assume $\zeta_6=e^{\frac{2\pi i~~~}{6~}}$, notice $({\color{red}4},6)=2$, we have $$z^3-1=(z-\underbrace{\zeta_6^{{\color{red}4}\cdot 1}}_{=\omega^2})(z-\underbrace{\zeta_6^{{\color{red}4}\cdot2}}_{=\omega})(z-\underbrace{\zeta_6^{{\color{red}4}\cdot3}}_{=\omega^3}).$$
Since $nq$ is a multiple of $2015$, we have $\xi^{\ell nq}=\left(\xi^{nq}\right)^\ell=1^\ell=1$ for every $\ell=1,2,...,m-1$, hence \begin{align*} \prod_{k=1}^{2015}\left(1+\xi^{nk}\right)&=\prod_{k=1}^{q}\left(1+\xi^{nk}\right)\cdot\prod_{k=q+1}^{2q}\left(1+\xi^{nk}\right)\cdots\prod_{k=(m-1)q+1}^{2015}\left(1+\xi^{nk}\right)\\ &=\prod_{\ell=0}^{m-1}\prod_{k=\ell q+1}^{(\ell+1)q}\left(1+\xi^{nk}\right)\\ &=\left(\prod_{k=1}^{q}\left(1+\xi^{nk}\right)\right)^m\\ &=(-f(-1))^m=(1-\underbrace{(-1)^q}_{\text {odd power}})^m=2^m. \end{align*}
From the claim, as $2015=5\cdot13\cdot31$, for every $d\mid 2015$, there are $\varphi\left(\frac{2015}{d}\right)$ numbers satisfying $(n,2015)=d$, hence the answer is the $2$ to the power of
$$\sum_{d\mid2015}d\cdot\varphi\left(\frac{2015}{d}\right)=~{\color{red}{13725}}.$$
PS: I mistook $403$ for prime..