11. (20 points) Let $n (n \geqslant 2)$ be a given positive integer. Find the value of $\prod_{k=1}^{n-1} \sin \frac{n(k-1)+k}{n(n+1)} \pi$.


11. Let $z_{k}=\cos \frac{2 k \pi}{n}+\mathrm{i} \sin \frac{2 k \pi}{n}(k=1,2, \cdots, n-1)$,
$$
\omega_{k}=\cos \frac{2 k \pi}{n+1}+\mathrm{i} \sin \frac{2 k \pi}{n+1}(k=1,2, \cdots, n) \text {. }
$$

Then the $n$ roots of $x^{n}-1=0$ are $1, z_{1}, z_{2}, \cdots, z_{n-1}$; the $n+1$ roots of $x^{n+1}-1=0$ are $1, \omega_{1}, \omega_{2}, \cdots, \omega_{n}$.
$$
\begin{array}{l}
\text { Hence }(x-1)\left(x^{n-1}+x^{n-2}+\cdots+1\right) \\
=(x-1)\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{n-1}\right) \\
\Rightarrow 1+x+\cdots+x^{n-2}+x^{n-1} \\
\quad=\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{n-1}\right) \\
\Rightarrow 1+x+\cdots+x^{n-1}+x^{n} \\
\quad=1+x\left(1+x+\cdots+x^{n-1}\right) \\
=1+x\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{n-1}\right) .
\end{array}
$$

On the other hand,
$$
\begin{array}{l}
(x-1)\left(x^{n}+x^{n-1}+\cdots+1\right) \\
=(x-1)\left(x-\omega_{1}\right)\left(x-\omega_{2}\right) \cdots\left(x-\omega_{n}\right) \\
\Rightarrow 1+x+\cdots+x^{n-1}+x^{n} \\
\quad=\left(x-\omega_{1}\right)\left(x-\omega_{2}\right) \cdots\left(x-\omega_{n}\right) \\
\Rightarrow 1+x\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{n-1}\right) \\
=\left(x-\omega_{1}\right)\left(x-\omega_{2}\right) \cdots\left(x-\omega_{n}\right) .
\end{array}
$$

Let $x=\omega_{1}$, we get
$$
\begin{array}{l}
1+\omega_{1}\left(\omega_{1}-z_{1}\right)\left(\omega_{1}-z_{2}\right) \cdots\left(\omega_{1}-z_{n-1}\right)=0 \\
\Rightarrow \omega_{1}\left(\omega_{1}-z_{1}\right)\left(\omega_{1}-z_{2}\right) \cdots\left(\omega_{1}-z_{n-1}\right)=-1 .
\end{array}
$$

Taking the modulus on both sides, we get
$$
\begin{array}{l}
1=\left|\omega_{1}\left(\omega_{1}-z_{1}\right)\left(\omega_{1}-z_{2}\right) \cdots\left(\omega_{1}-z_{n-1}\right)\right| \\
=\left|\omega_{1}\right| \prod_{k=1}^{n-1}\left|\omega_{1}-z_{k}\right|=\prod_{k=1}^{n-1}\left|\omega_{1}-z_{k}\right| . \\
\text { Also, }\left|\omega_{1}-z_{k}\right|^{2} \\
=\left(\cos \frac{2 k \pi}{n}-\cos \frac{2 \pi}{n+1}\right)^{2}+\left(\sin \frac{2 k \pi}{n}-\sin \frac{2 \pi}{n+1}\right)^{2} \\
=2-2 \cos \left(\frac{2 k \pi}{n}-\frac{2 \pi}{n+1}\right) \\
=4 \sin ^{2} \frac{n(k-1)+k}{n(n+1)} \pi \\
\Rightarrow\left|\omega_{1}-z_{k}\right|=2 \sin \frac{n(k-1)+k}{n(n+1)} \pi .
\end{array}
$$

Thus, equation (1) becomes
$$
\begin{array}{l}
1=\prod_{k=1}^{n-1}\left[2 \sin \frac{n(k-1)+k}{n(n+1)} \pi\right] \\
=2^{n-1} \prod_{k=1}^{n-1} \sin \frac{n(k-1)+k}{n(n+1)} \pi . \\
\text { Hence } \prod_{k=1}^{n-1} \sin \frac{n(k-1)+k}{n(n+1)} \pi=\frac{1}{2^{n-1}} .
\end{array}
$$