It is too long for a comment, so I post it as answer.
I will present two solutions.
First solution:
The approach proposed by Max Alekseyev is correct.
Let us prove
$\dfrac{k}{2^{2k-2}}=(\prod_{j=1}^k \sin \frac{\pi j}{2k})^2$, for $k \in \mathbb{N}, k > 0$
Let us go step by step.
Step 1. Given $k \in \mathbb{N}, k > 0$, let us consider the Chebyshev polynomial of the second kind $U_{2k-1}$.
We have, by definition, that
$$U_{2k-1}(\cos \theta)\sin \theta= \sin 2k \theta \tag{1}$$
and its explicit form is:
$$U_{2k-1}(x) = \sum_{t=0}^{k-1} (-1)^t \binom{{2k-1}-t}{t}~(2x)^{{2k-1}-2t} \tag{2}$$
Step 2. From $(2)$, we see that:
- the degree of $U_{2k-1}$ is $2k-1$;
- the leading coefficient of $U_{2k-1}$ is $2^{2k-1}$
- the term of lowest degree is $(-1)^{k-1}\binom{k}{k-1}~(2x)$, that is $(-1)^{k-1}2kx$
Step 3. Let us look at the roots of $U_{2k-1}$. From $(1)$,
taking $\theta = \frac{r \pi}{2k}$ where $r= 1, \dots, 2k-1$, we have
$$U_{2k-1}\left (\cos \frac{r \pi}{2k} \right )\sin\frac{r \pi}{2k}= \sin 2k \frac{r \pi}{2k} = \sin r \pi =0$$
Since for each $r= 1, \dots, 2k-1$, $\sin\frac{r \pi}{2k}\ne 0$, we have that, for each $r= 1, \dots, 2k-1$,
$$ U_{2k-1}\left (\cos \frac{r \pi}{2k} \right ) = 0$$
So we have that $U_{2k-1}$ has $2k-1$ distinct roots and they are $\cos \frac{r \pi}{2k}$ for $r= 1, \dots, 2k-1$.
Now, note that for each $r= 1, \dots, 2k-1$,
$$ \cos \frac{r \pi}{2k} = \cos \left ( \frac{\pi}{2}+ \frac{(r -k) \pi}{2k} \right ) = -\sin \frac{(r -k) \pi}{2k}$$
So the roots of $ U_{2k-1}$ are $-\sin \frac{(r -k) \pi}{2k}$ for $r= 1, \dots, 2k-1$. In other words (taking $s = r-k$), the roots of $ U_{2k-1}$ are $-\sin \frac{s \pi}{2k}$ for $s= -(k-1), \dots, k-1$.
It means that the roots of $ U_{2k-1}$ are $0$ and $\pm \sin \frac{s \pi}{2k}$ for $s= 1, \dots, k-1$.
Step 4. Since the term of lowest degree in $ U_{2k-1}$ is $(-1)^{k-1}2kx$ (or, equivalently, $0$ is a simples root of $ U_{2k-1}$), we can divide $ U_{2k-1}$ by $x$. Let $P$ be the polynomial obtained by such division. It is immediate that
- $P$ has degree $2k-2$;
- the roots of $P$ are $\pm \sin \frac{s \pi}{2k}$ for $s= 1, \dots, k-1$;
- the leading coefficient of $P$ is $2^{2k-1}$
- the term of lowest degree is $(-1)^{k-1}2k$
It follows immediately that
$$(-1)^{k-1}\left (\prod_{s=1}^{k-1} \sin \frac{s\pi }{2k} \right)\left (\prod_{s=1}^{k-1} \sin \frac{s\pi }{2k} \right)=(-1)^{(2k-2)} \frac{(-1)^{k-1}2k}{2^{2k-1}}$$
that is
$$\left (\prod_{s=1}^{k-1} \sin \frac{s\pi }{2k} \right)^2=\frac{k}{2^{2k-2}}$$
Since $\sin \frac{k\pi }{2k}= \sin \frac{\pi }{2} =1$, we can write:
$$\left (\prod_{s=1}^k \sin \frac{s\pi }{2k} \right)^2=\frac{k}{2^{2k-2}}$$
$\square$
Second solution: Here is another way to prove, using complex numbers, that
$\dfrac{k}{2^{2k-2}}=(\prod_{j=1}^k \sin \frac{\pi j}{2k})^2$, for $k \in \mathbb{N}, k > 0$
Proof: For $k=1$, it is immediate that
$$\dfrac{k}{2^{2k-2}}= 1 = \left (\sin \frac{\pi }{2} \right)^2 =\left (\prod_{j=1}^k \sin \frac{\pi j}{2k} \right )^2$$
Now, note that for $m \in \Bbb N$, $m \geqslant 1$
$$\frac{x^{m} - 1}{x-1} = \prod_{j=1}^{m-1} \left (x-e^\frac{2j\pi i} {m} \right )$$
On the other hand,
$$\frac{x^{m} - 1}{x-1} = x^{m-1}+ x^{m-2}+ \cdots+ x +1$$
So
$$ x^{m-1}+ x^{m-2}+ \cdots+ x +1 = \prod_{j=1}^{m-1} \left (x-e^\frac{2j\pi i} {m} \right )$$
Making $x=1$, we get
$$ m = \prod_{j=1}^{m-1} \left (1-e^\frac{2j\pi i} {m} \right )$$
So, for any $k \in \Bbb N$, $k \geqslant 2$, we have (making $m=2k$),
$$ 2k = \prod_{j=1}^{2k-1} \left (1-e^\frac{j\pi i} {k} \right )$$
Now, note that, for all $j \in \{1, \dots ,k-1\}$, $1-e^\frac{j\pi i} {k}$ is the conjugate of $1-e^\frac{(2k-j)\pi i} {k}$. So we have
\begin{align*} 2k &= \prod_{j=1}^{2k-1} \left (1-e^\frac{j\pi i} {k} \right )= \\
&= \left [\prod_{j=1}^{k-1} \left (1-e^\frac{j\pi i} {k} \right ) \right ] \left (1-e^\frac{k\pi i} {k} \right ) \left [\prod_{j=k+1}^{2k-1} \left (1-e^\frac{j\pi i} {k} \right ) \right ] = \\
&= 2 \left [\prod_{j=1}^{k-1} \left (1-e^\frac{j\pi i} {k} \right ) \right ] \left [\prod_{j=k-1}^{1} \left (1-e^\frac{(2k-j)\pi i} {k} \right ) \right ] = \\
&= 2 \prod_{j=1}^{k-1} \left |1-e^\frac{j\pi i} {k} \right |^2 =\\
&= 2 \prod_{j=1}^{k-1} \left[ \left (1- \cos \frac{j\pi } {k} \right )^2 + \left (\sin \frac{j\pi } {k} \right )^2 \right] =\\
&= 2 \prod_{j=1}^{k-1} \left[ 2 \left (1- \cos \frac{j\pi } {k} \right ) \right] =\\
&= 2^k \prod_{j=1}^{k-1} \left (1- \cos \frac{j\pi } {k} \right ) = \\
&= 2^k \prod_{j=1}^{k-1} \left[ 2 \left (\sin \frac{j\pi } {2k} \right )^2 \right] =\\
&= 2^{2k-1} \prod_{j=1}^{k-1} \left (\sin \frac{j\pi } {2k} \right )^2
\end{align*}
note that we used the fact that $1 -\cos \theta = 2 \left(\sin \frac{\theta}{2} \right)^2$.
So, we have
$$ \frac{k}{2^{2k-2}}=\left (\prod_{j=1}^{k-1} \sin \frac{j\pi }{2k} \right )^2 $$
Since $ \sin \frac{k\pi }{2k} = \sin \frac{\pi }{2}=1$, we can write
$$ \frac{k}{2^{2k-2}}=\left (\prod_{j=1}^{k} \sin \frac{j\pi }{2k} \right )^2 $$