Productive polygons (17/09/21 Riddler Classic)
The remarkable result is that, for a regular $N$-gon with side lengths $2$, the diagonal-fragment product is simply $$\begin{cases}2, &N\text{ even}\\\sqrt{N}, &N\text{ odd}\end{cases}$$
$N$ is even
Consider the $N$-gon where $N=2m$. Given the side-length of 2, the "radius" (centre-to-vertex) of the polygon is given by $$r=\csc\left(\frac{\pi}{2m}\right)$$
We now carve up the long diagonal as specificed.
Shaded red is the $k$th piece of the diagonal counting from the top; in this example figure, $k=3$ where $m=7$. In general, the diagonal fragment's length $L_k$ is given by$$L_k=r\left(\cos\left(\frac{(k-1)\pi}{m}\right)-\cos\left(\frac{k\pi}{m}\right)\right)$$
Rearranging this using some trig identities:
$$\begin{align*}\frac{L_k}{r}&=\cos\left(\frac{k\pi}{m}\right)\cos\left(\frac{\pi}{m}\right)+\sin\left(\frac{k\pi}{m}\right)\sin\left(\frac{\pi}{m}\right)-\cos\left(\frac{k\pi}{m}\right)\\&=\cos\left(\frac{k\pi}{m}\right)\left[1-2\sin^2\left(\frac{\pi}{2m}\right)\right]+2\sin\left(\frac{k\pi}{m}\right)\sin\left(\frac{\pi}{2m}\right)\cos\left(\frac{\pi}{2m}\right)-\cos\left(\frac{k\pi}{m}\right)\\&=2\sin\left(\frac{\pi}{2m}\right)\left[\sin\left(\frac{k\pi}{m}\right)\cos\left(\frac{\pi}{2m}\right)-\sin\left(\frac{\pi}{2m}\right)\cos\left(\frac{k\pi}{m}\right)\right]\\&=2\sin\left(\frac{\pi}{2m}\right)\sin\left(\frac{(2k-1)\pi}{2m}\right)\\\therefore L_k&=2\sin\left(\frac{(2k-1)\pi}{2m}\right)\end{align*}$$
And so, taking the product of $L_k$ over $k\in[1,m]$:
$$\begin{align*}\color{red}{\boxed{\prod_{k=1}^mL_k}}&=2^m\prod_{k=1}^m \sin\left(\frac{(2k-1)\pi}{2m}\right)\\&=\color{red}{\boxed{2}}\end{align*}$$
where we used the really cool identity $$\prod_{k=1}^m \sin\left(\frac{(2k-1)\pi}{2m}\right)=\frac{1}{2^{m-1}}$$
Proof of the cool identity
$$\begin{align*}\prod_{k=1}^m \sin\left(\frac{(2k-1)\pi}{2m}\right)&=\frac{1}{(2i)^m}\prod_{k=1}^m \left[\exp\left(\frac{i(2k-1)\pi}{2m}\right)-\exp\left(-\frac{i(2k-1)\pi}{2m}\right)\right]\\&=\frac{1}{(2i)^m}\prod_{k=1}^m \left[\exp\left(\frac{i(2k-1)\pi}{2m}\right)\left[1-\exp\left(-\frac{i(2k-1)\pi}{m}\right)\right]\right]\\&=\frac{1}{(2i)^m}\cdot\exp\left(\frac{i\pi}{2m}\sum_{k=1}^m(2k-1)\right)\prod_{k=1}^m\left[1-\exp\left(-\frac{i(2k-1)\pi}{m}\right)\right]\\&=\frac{1}{(2i)^m}\cdot\exp\left(\frac{i\pi}{2m}m^2\right)\prod_{k=1}^m\left[1-\exp\left(-\frac{i(2k-1)\pi}{m}\right)\right]\\&=\frac{1}{(2i)^m}\cdot\exp\left(\frac{im\pi}{2}\right)\cdot f(1)\end{align*}$$
where $$f(z)\;\dot{=}\;\prod_{k=1}^m\left[z-\exp\left(-\frac{i(2k-1)\pi}{m}\right)\right]$$
Specifically, note that $\exp\left(-\frac{i(2k-1)\pi}{m}\right)$ is the general form for the zeros of the polynomial $z^m+1=0$. This means that $f(z)=z^m+1$. Continuing the working from above, we complete the proof of the identity:
$$\begin{align*}\prod_{k=1}^m \sin\left(\frac{(2k-1)\pi}{2m}\right)&=\frac{1}{(2i)^m}\cdot\exp\left(\frac{im\pi}{2}\right)\cdot f(1)\\&=\frac{1}{(2i)^m}\cdot i^m\cdot 2\\&=\frac{1}{2^{m-1}}\end{align*}$$
$N$ is odd
Consider the $N$-gon where $N=2m+1$. Similarly to the even-$N$ case, the $m$ diagonal fragments (counting from the apex) have lengths given by $$L_k=2\sin\left(\frac{(2k-1)\pi}{2m+1}\right)$$
And so the diagonal-fragment product is $$\begin{align*}\color{red}{\boxed{\prod_{k=1}^mL_k}}&=2^m\prod_{k=1}^m\sin\left(\frac{(2k-1)\pi}{2m+1}\right)\\&=\sqrt{2m+1}\\&=\color{red}{\boxed{\sqrt{N}}}\end{align*}$$
where we've used the even-cooler identity $$\prod_{k=1}^m\sin\left(\frac{(2k-1)\pi}{2m+1}\right)=\frac{\sqrt{2m+1}}{2^m}$$
Proof of the even-cooler identity
In a similar manner to how we approached the identity from the even-$N$ case:
$$\begin{align*}\prod_{k=1}^m\sin\left(\frac{(2k-1)\pi}{2m+1}\right)&=\frac{1}{(2i)^m}\prod_{k=1}^m\left[\exp\left(\frac{i(2k-1)\pi}{2m+1}\right)\left[1-\exp\left(-\frac{2i(2k-1)\pi}{2m+1}\right)\right]\right]\\&=\frac{1}{(2i)^m}\prod_{k=1}^m\exp\left(\frac{i(2k-1)\pi}{2m+1}\right)\prod_{k=1}^m\left[1-\exp\left(-\frac{2i(2k-1)\pi}{2m+1}\right)\right]\\&=\frac{1}{(2i)^m}\exp\left(\frac{i\pi m^2}{2m+1}\right)\prod_{k=1}^m\left[1-\exp\left(-\frac{2i(2k-1)\pi}{2m+1}\right)\right]\tag{1}\\\end{align*}$$
Once again, we want to reduce the product to a polynomial. In this case, we consider the polynomial $(z^{2m+1}-1)$, which has $2m+1$ zeros. Specifically, the zeros are exponentials of the integer multiples of $\frac{-2i\pi}{2m+1}$, where these integer multiples are in $[0,2m]$. Hence we can factorise $(z^{2m+1}-1)$ as $$\begin{align*}(z^{2m+1}-1)&=(z-1)\cdot\prod_{k=1}^m\left[z-\exp\left(-(2k-1)\frac{2i\pi}{2m+1}\right)\right]\cdot\prod_{k=1}^m\left[z-\exp\left(-2k\frac{2i\pi}{2m+1}\right)\right]\\&=(z-1)\cdot O_m(z)\cdot E_m(z)\end{align*}$$where$$O_m(z)\;\dot{=}\;\prod_{k=1}^m\left[z-\exp\left(-(2k-1)\frac{2i\pi}{2m+1}\right)\right]$$are the polynomial factors resulting from odd-integer multiples of $\frac{-2i\pi}{2m+1}$, while$$E_m(z)\;\dot{=}\;\prod_{k=1}^m\left[z-\exp\left(-2k\frac{2i\pi}{2m+1}\right)\right]$$are the nonzero even-integer multiples.
One crucial fact about these zeros of unity is that the nonzero odd- and even-multiples of $\frac{-2i\pi}{2m+1}$ can be paired up as complex conjugates. Specifically, for some $k\in[1,2m]$, the number $(2m+1-k)$ is of the opposite parity. These two multiples result in zeros that are conjugates, like so: $$\exp\left(-k\frac{2i\pi}{2m+1}\right)^\ast=\exp\left(-(2m+1-k)\frac{2i\pi}{2m+1}\right)$$
This means that for $z\in\mathbb{R}$, the products $O_m(z)$ and $E_m(z)$ are also conjugates since they contain factors that are odd-even conjugate pairs. $$O_m(z)=E_m(z)^*$$
And so for $z\in\mathbb{R}$, $$\begin{align*}(z^{2m+1}-1)&=(z-1)O_m(z)E_m(z)\\&=(z-1)O_m(z)O_m(z)^*\\&=(z-1)|O_m(z)|^2\\\therefore |O_m(z)|&=\sqrt{\frac{z^{2m+1}-1}{z-1}}\\&=\sqrt{\sum_{k=0}^{2m}z^k}\end{align*}$$
With all this information, we return finally to the actual main working out, which we left off at $(1)$.
$$\begin{align*}\prod_{k=1}^m\sin\left(\frac{(2k-1)\pi}{2m+1}\right)&=\frac{1}{(2i)^m}\exp\left(\frac{i\pi m^2}{2m+1}\right)\cdot\prod_{k=1}^m\left[1-\exp\left(-\frac{2i(2k-1)\pi}{2m+1}\right)\right]\\&=\frac{1}{(2i)^m}\exp\left(\frac{i\pi m^2}{2m+1}\right)\cdot O_m(1)\end{align*}$$
Now, the product of sines on the LHS will clearly be a positive real number. Hence, to eliminate any complex stuff from the RHS, we can simply equate it to its complex modulus.
$$\begin{align*}\prod_{k=1}^m\sin\left(\frac{(2k-1)\pi}{2m+1}\right)&=\Big|\prod_{k=1}^m\sin\left(\frac{(2k-1)\pi}{2m+1}\right)\Big|\\&=\frac{1}{2^m}\cdot|O_m(1)|\\&=\frac{1}{2^m}\cdot\sqrt{\sum_{k=0}^{2m}z^k}\Big|_{z=1}\\&=\frac{\sqrt{2m+1}}{2^m}\end{align*}$$
which completes the proof of the identity.