Product of Sines of Fractions of Pi

Theorem
Let $m \in \Z$ such that $m > 1$.

Then:
 * $\ds \prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} m = \frac m {2^{m - 1} }$

Proof
Consider the equation:
 * $z^m - 1 = 0$

whose solutions are the complex roots of unity:
 * $1, e^{2 \pi i / m}, e^{4 \pi i / m}, e^{6 \pi i / m}, \ldots, e^{2 \paren {m - 1} \pi i / m}$

Then:

Also presented as
Some sources prefer to report this result as:
 * $\ds \prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} m = \frac {2 m} {2^m}$

on the grounds that it may be easier to remember.