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.

I like the answer by @Steven Gregory because of the way the dominos fall and it seems the only one presented that a precalculus student could hope to find. Using the reflection and multiplication formulas for the gamma function a fairly compact proof is possible.
$$\begin{align}\prod_{r=1}^7\cos\frac{r\pi}{15} & = \prod_{r=1}^7\sin\left(\frac{\pi}2-\frac{r\pi}{15}\right) = \prod_{r=0}^6\sin\left(\frac{\pi}2-\frac{7\pi}{15}+\frac{r\pi}{15}\right) \\
& = \prod_{r=0}^6\sin\left(\frac{\pi}{30}+\frac{r\pi}{15}\right) = \prod_{r=0}^6\frac{\pi}{\Gamma\left(\frac1{30}+\frac{r}{15}\right)\Gamma\left(\frac{29}{30}-\frac{r}{15}\right)} & \tag{1} \\
& = \frac{\pi^7}{\prod_{r=0}^6\Gamma\left(\frac1{30}+\frac{r}{15}\right)\prod_{s=8}^{14}\Gamma\left(\frac1{30}+\frac{s}{15}\right)} = \frac{\pi^7\,\Gamma\left(\frac1{30}+\frac7{15}\right)}{\prod_{r=0}^{14}\Gamma\left(\frac1{30}+\frac{r}{15}\right)} & \tag{2} \\
& = \frac{\pi^7\,\Gamma\left(\frac12\right)}{(2\pi)^{(15-1)/2}15^{\frac12-15\left(\frac1{30}\right)}\Gamma\left(15\left(\frac1{30}\right)\right)} & \tag{3} \\
& = \frac1{2^7}
\end{align}$$
$(1)$ Using the reflection formula for the gamma function: $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$.
$(2)$ Reordering one product and multiplying and dividing by $\Gamma\left(\frac1{30}+\frac7{15}\right)$.
$(3)$ Using the multiplication formula for the gamma function: $$\prod_{k=0}^{n-1}\Gamma\left(x+\frac{k}n\right)=(2\pi)^{(n-1)/2}n^{\frac12-nx}\Gamma(nx)$$
