<s>[INST] Closed form for $\prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$ Question: I am hoping to analytically continue the function:
$F(n) = \prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$
to $n=1/2$. My understanding is that means I will need a closed form for this product. What is a closed form of the above product? (Or are there other methods for performing the analytic continuation without having a closed form?)
Attempt: A similar product satisfies the identity
$\prod_{k=1}^{n-1} \sin(\frac{\pi k}{n}) = n 2^{1-n}$.
A proof of the above identity is given in another post (Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$). I have gone through the proofs given there and tried to extend the techniques used to my case, but no luck so far. 
Any help is appreciated - a full solution, or just suggestions of theorems or related identities which may be of use.
 [/INST] We break up the product in a different way (analogous to changing the order of products):
$$\begin{align}
F(n) &= \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \sin{\left(\frac{3\pi}{n}\right)} \\
&\quad \quad \vdots \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}.
\end{align}$$
(Note there are $n-1$ of the first term, $n-2$ of the second, down to one of the last). Now, consider $F(n)^2$: it is sufficient to do this and square root afterwards since every term in the finite product is positive, so the whole thing must be positive. Again, we reorder the terms:
$$\begin{align}
F(n)^2 &= \sin{\left(\frac{(n-1)\pi}{n}\right)} \sin{\left(\frac{(n-2)\pi}{n}\right)} \dotsm \sin{\left(\frac{\pi}{n}\right)} \\ 
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \times \sin{\left(\frac{(n-2)\pi}{n}\right)} \dotsm \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \quad \vdots \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \times \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}.
\end{align}$$
(so basically I've taken the triangle of terms, flipped it over and made a rectangle).
Now, $\sin{\left(\frac{(n-k)\pi}{n}\right)} = \sin{\left(\frac{k\pi}{n}\right)}$, so we can change the terms in the upper-right triangle as follows:
$$\begin{align}
F(n)^2 &= \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-1)\pi}{n}\right)} \\ 
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \times \sin{\left(\frac{2\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-1)\pi}{n}\right)} \\
&\quad \quad \vdots \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \times \sin{\left(\frac{(n-1)\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}.
\end{align}$$
But of course this is just $n-1+1=n$ copies of $\prod_{k=1}^{n-1} \sin{\left( \frac{k\pi}{n} \right)} = n/2^{n-1}$, so we find
$$ F(n) = \sqrt{\frac{n^n}{2^{n(n-1)}}} = \frac{n^{n/2}}{2^{n(n-1)/2}}, $$
as conjectured by Claude Leibovici.
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