I was attempting a problem about the expansion of n! using sine function.
The photograph contains the exact statement of the problem and my attempted solution.
I am looking for the completion of my attempt or a different solution to the problem.
The problem is:
$$u! = 2^{u(u-1)/2}\prod_{m=2}^u\prod_{k=1}^{m-1}\sin\left(\frac{k\pi}m\right)$$
We can write $$\sin\frac{k\pi}{m}=\frac{e^{k\pi i/m}-e^{-k\pi i/m}}{2i} =-\frac{e^{-k\pi i/m}}{2i}(z-e^{2k\pi i/m})\quad\hbox{with}\quad z=1\ .$$ If we multiply all these terms for $k$ from $1$ to $m-1$, the exponentials become $$e^{-(1+2+\cdots+(m-1))\pi i/m}=e^{-(m-1)\pi i/2}=(-i)^{m-1}$$ and the terms with $z$s become $$\frac{z^m-1}{z-1}\ .$$ To check the last statement note that the numerator is the product of all factors $z-e^{2k\pi i/m}$ with $k$ from $0$ to $m-1$, and the denominator then cancels out the term with $k=0$. Putting all this together we get $$\prod_{k=1}^{m-1}\sin\frac{k\pi}{m} =\frac{1}{2^{m-1}}\lim_{z\to1}\frac{z^m-1}{z-1} =\frac{m}{2^{m-1}}$$ by L'Hopital's Rule. Therefore $$\prod_{m=2}^u\prod_{k=1}^{m-1}\sin\frac{k\pi}{m} =\frac22\times\frac34\times\cdots\times\frac{u}{2^{u-1}} =\frac{u!}{2^{1+2+\cdots+(u-1)}} =\frac{u!}{2^{u(u-1)/2}}\ .$$
Use the result that $$2^{n-1} \prod_{k=0}^{n-1} \sin(x+\frac{k\pi}{n})=\sin nx$$ Now divide both sides by $\sin x$ and take limit as $x \to 0$ You will get $$2^{n-1} \prod_{k=1}^{n-1} \sin(k\pi/n)=n$$ which implies $$\prod_{k=1}^{m-1} \sin(k \pi/m)=\frac{2n}{2^n}$$ Now it is fairly easy to proceed.
First, we will prove
$$u=2^{u-1}\prod_{k=1}^{u-1}\sin\left(\frac{k\pi}u\right)$$
We will do this by a strange form of induction; assume it's true for $u$, and prove that it's true for $2u$ and $2u+1$. The expression is true for $u=1$. Now assume it's true for $u$, then:
\begin{align} 2^{2u-1}\prod_{k=1}^{2u-1}\sin\left(\frac{k\pi}{2u}\right)&=2^{2u-1}\sin\left(\frac{u\pi}{2u}\right)\prod_{k=1}^{u-1}\sin\left(\frac{k\pi}{2u}\right)\sin\left(\pi-\frac{k\pi}{2u}\right)\\ &=2^{2u-1}\sin\left(\frac\pi2\right)\prod_{k=1}^{u-1}\sin\left(\frac{k\pi}{2u}\right)\sin\left(\frac{k\pi}{2u}\right)\\ &=2^{2u-1}\prod_{k=1}^{u-1}\sin\left(\frac{k\pi}{2u}\right)^2\\ &=2^{2u-1}\left(\prod_{k=1}^{u-1}\sin\left(\frac{k\pi}{2u}\right)\right)^2\\ &=2^{2u-1}\prod_{k=1}^{u-1}\sin\left(\frac{k\pi}{2u}\right)\prod_{k=1}^{u-1}\cos\left(\frac{k\pi}{2u}\right)\\ &=2^{u}\prod_{k=1}^{u-1}2\sin\left(\frac{k\pi}{2u}\right)\cos\left(\frac{k\pi}{2u}\right)\\ &=2\cdot2^{u-1}\prod_{k=1}^{u-1}\sin\left(\frac{k\pi}u\right)\\ &=2u \end{align}
And we can do the same thing to prove it for $2u+1$ (which I will not do here because it's practically the same and will make this post longer than needed), proving it for all $u$.
Trivially, the expression is true for $u=0$ and $u=1$. Now assume that it's true for $u$. Then
\begin{align} 2^{(u+1)u/2}\prod_{m=2}^{u+1}\prod_{k=1}^{m-1}\sin\left(\frac{k\pi}m\right)&=2^{u(u-1)/2+u}\prod_{m=2}^{u+1}\prod_{k=1}^{m-1}\sin\left(\frac{k\pi}m\right)\\ &=2^u2^{u(u-1)/2}\prod_{m=2}^u\prod_{k=1}^{m-1}\sin\left(\frac{k\pi}m\right)\cdot\prod_{k=1}^u\sin\left(\frac{k\pi}{u+1}\right)\\ &=2^u\prod_{k=1}^u\sin\left(\frac{k\pi}{u+1}\right)\cdot\left(2^{u(u-1)/2}\prod_{m=2}^u\prod_{k=1}^{m-1}\sin\left(\frac{k\pi}m\right)\right)\\ &=2^u\prod_{k=1}^u\sin\left(\frac{k\pi}{u+1}\right)\cdot u!\\ &=(u+1)\cdot u!\\ &=(u+1)! \end{align}
And that proves the statement.