Integral exercise asked in many math groups for the product of cosines

Exercise:

Compute, $\int_{0}^{2\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}}$

Solution: Let $f\left( x \right)=\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)}$

We know that $\cos x$ is a periodic function of period $w=2p$

i.e $f\left( x \right)=f\left( w-x \right)$ so $\int_{0}^{2p}{f\left( x \right)dx}=2\int_{0}^{p}{f\left( x \right)dx}$

$f\left( 2\pi -x \right)=\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( n\left( 2\pi -x \right) \right)}=\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( 2n\pi -nx \right)=\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( -nx \right)=f\left( x \right)}}$

So $\int_{0}^{2\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}=2\int_{0}^{\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}}}$

But $\int_{0}^{\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}=\int_{0}^{\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( \pi -nx \right)dx}=-\int_{0}^{\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}}}}$

$\Leftrightarrow \int_{0}^{2\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}=-2\int_{0}^{\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}}}=-\int_{0}^{2\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}}$

$\Leftrightarrow \int_{0}^{2\pi }{\prod\limits_{n=1}^{2016}{{{\cos }^{n}}\left( nx \right)dx}}=0$