Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $ 
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$

I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help will be appreciated.
Thanks.

Since an elegant solution has already been provided, I will go for an overkill.
From the Fourier cosine series of $\log\cos x$ we have:
$$ \log\cos x = -\log 2-\sum_{n\geq 1}^{+\infty}\frac{(-1)^n\cos(2n x)}{n}\tag{1} $$
but for any $n\geq 1$ we have:
$$ 15\nmid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = -\frac{1}{2},\quad 15\mid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = 7\tag{2}$$
so:
$$ \sum_{k=1}^{7}\log\cos\frac{k\pi}{15} = -7\log 2+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n}-\frac{15}{2}\sum_{n\geq 1}\frac{(-1)^n}{15n}=-7\log 2\tag{3}$$
and by exponentiating the previous line:

$$ \prod_{k=1}^{7}\cos\left(\frac{\pi k}{15}\right) = \color{red}{\frac{1}{2^7}}.\tag{4}$$

