Wronskian of functions $\sin(nx), n=1,2,...,k$. Is it true that the Wronskian of the functions $\sin(nx), n=1,...,k$ is equal to $c(\sin(x))^p$ where $c$ is a constant and $p=1+2+...+k=k(k+1)/2$?
That is true for $k=1,2,3,4,5$. If it is true, how to find the constant $c=c(k)$?

Let $W(f_1, \ldots, f_n)$ denote the Wronskian determinant of the functions $f_1, \ldots, f_n$. We can show that

$$
W (\sin(x), \sin(2x), \ldots, \sin(nx)) = 
1!2! \cdots (n-1)! (-2)^{n(n-1)/2} \sin(x)^{n(n+1)/2} \, .
$$

For example,
$$
\begin{align}
 W (\sin(x), \sin(2x)) &= -2 (\sin(x))^3 \, .\\
 W (\sin(x), \sin(2x), \sin(3x)) &= -16 (\sin(x))^{6} \, ,\\
 W (\sin(x), \ldots, \sin(4x)) &= 768 (\sin(x))^{10} \, .
\end{align}  
$$
The proof uses that
$$
 \sin(k x) =  U_{k-1}(\cos (x)) \sin(x)
$$
where $U_k$ are the Chebyshev polynomials of the second kind, and two identities for Wronskians: A “product rule”
$$
W(h f_1, \ldots, h f_n) = h^n \cdot W(f_1, \ldots, f_n)
$$
which is a consequence of the Leibniz rule for the $n$th derivative of a product (see also Why does the Wronskian satisfy $W(yy_1,\ldots,yy_n)=y^n W(y_1,\ldots,y_n)$?), and a “chain rule”
$$
 W(f_1 \circ g, \ldots, f_n \circ g)(x) = W(f_1, \ldots f_n)(g(x)) \cdot (g'(x))^{n(n-1)/2} \, .
$$
which is a consequence of Faà di Bruno's formula for the $n$th derivative of a composite function (compare also About a chain rule for Wronskians).
Now we can argue as follows:
$$
\begin{align}
&W (\sin(x), \sin(2x), \ldots, \sin(nx)) \\
&\quad = W(U_0(\cos(x))\sin(x), U_1(\cos (x)) \sin(x), \ldots,  U_{n-1}(\cos (x)) \sin(x)) \\
&\quad = (\sin(x))^n W(U_0(\cos (x)), U_1(\cos (x)), \ldots,  U_{n-1}(\cos (x))) \, \\
&\quad = (\sin(x))^n W(U_0(t), U_1(t), \ldots,  U_{n-1}(t)) |_{t=\cos(x)} (-\sin(x))^{n(n-1)/2} \, .
\end{align}
$$
Each $U_k$ is a polynomial of degree $k$ with the leading coefficient $2^k$, so that $W(U_0, U_1, \ldots, U_{n-1})$ is the determinant of a triangular matrix with the entries $U_k^{(k)}(t) = k!2^k$, $k=0, \ldots, n-1$ on the diagonal. It follows that
$$
 W (\sin(x), \sin(2x), \ldots, \sin(nx)) = (\sin(x))^n \cdot (-\sin(x))^{n(n-1)/2}
\cdot \prod_{k=0}^{n-1} k!2^k
$$
and that is the claimed formula.
