Product of sine For which $n\in \mathbb{N}$, can we find (reps. find explicitly) $n+1$ integers $0 < k_1 < k_2 <\cdots < k_n < q<2^{2n}$
such that 
$$\prod_{i=1}^{n} \sin\left(\frac{k_i \pi}{q} \right) =\frac{1}{2^n} $$
P.S.: $n=2$ is obvious answer, $n=6 $ is less obvious but for instance we have $k_1 = 1$, $k_2 = 67$, $k_3 = 69$, $k_4 = 73$, $k_5 = 81$, $k_6 = 97$, and $q=130$.

Consider the identity (as quoted by drvitek):
$$\prod_{k=1}^{n} \sin \left(\frac{(2k-1) \pi}{2n}\right) = \frac{2}{2^n},$$
This is completely correct but doesn't quite answer the question because the RHS is
not $1/2^n$ $\text{---}$ this is an issue related to the fact that $\zeta - \zeta^{-1}$ is not
a unit if $\zeta$ is a root of unity of prime power order.
Replace $n$ by $3n$, and take the ratio of the corresponding
products. Then one finds that
$$\prod_{(k,6) = 1}^{k < 6m} \sin \left(\frac{k \pi}{6m}\right)=
\sin \left(\frac{\pi}{6m}\right)
\sin \left(\frac{5 \pi}{6m}\right)
\sin \left(\frac{7 \pi}{6m}\right) \cdots
\sin \left(\frac{(6m-1) \pi}{6m}\right) = \frac{1}{2^{2m}}.$$
This provides the identity you request for $n = 2m$, since
$q = 6m < 2^{4m} = 2^{2n}$ is true for all $m \ge 1$.
There are other "obvious" identities that can be written down, but they tend to have length
$\phi(r)$ for some integer $r$, and $\phi(r)$ is always even (if $r > 2$).
For odd $n$, note the "exotic" identity:
$$\sin \left(\frac{2 \pi}{42}\right) 
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right) = \frac{1}{8}.$$
Since $42 < 64$, this is an identity of the required form for $n = 3$.
On the other hand, none of the rational numbers
$1/21$, $5/14$, $8/21$ can be written in the form $k/6m$
 where $(k,6) = 1$.
Hence
$$\sin \left(\frac{2 \pi}{42}\right)
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right) 
\prod_{(k,6) = 1}^{k < 6m} \sin \left(\frac{k \pi}{6m}\right) = \frac{1}{2^{2m+3}},$$
when written under the common denominator $q = \mathrm{lcm}(42,6m)$, consists of distinct fractional multiples $k_i/q$ of $\pi$ with $0 < k_i < q$, and is thus
 an identity of the required form for $n = 2m + 3$, after checking that
$$q = \mathrm{lcm}(42,6m) \le 42m \le 2^{4m+6} = 2^{2n}.$$
Thus the answer to your question is that such an identity holds for all $n > 1$.
(It trivially does not hold for $n = 1$.)
The first few identities constructed in this way are:
$$\sin \left(\frac{\pi}{6}\right) \sin \left(\frac{5 \pi}{6}\right) = 
\frac{1}{4},$$
$$\sin \left(\frac{2 \pi}{42}\right)
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right) = \frac{1}{8},$$
$$\sin \left(\frac{\pi}{12}\right) \sin \left(\frac{5 \pi}{12}\right)      
\sin \left(\frac{7 \pi}{12}\right) \sin \left(\frac{11 \pi}{12}\right) =     
\frac{1}{16},$$
$$\sin \left(\frac{2 \pi}{42}\right) \sin \left(\frac{7 \pi}{42}\right) 
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right)
\sin \left(\frac{35 \pi}{42}\right)  = \frac{1}{32},$$
&. &.
