Prove the formula for Jacobian determinant for this coordinate transform for general $n$ Let $n$ be some integer $n \geq 2$, $(x_1, \cdots, x_n)$ are the old coordinates and $(r, \phi_1, \phi_2, \cdots, \phi_{n-1})$ are some new coordinates. I want to transform $(x_1, \cdots, x_n)$ to $(r, \phi_1, \cdots, \phi_{n-1})$ in the following way
$$
 \begin{aligned}
  x_1 &= r \cos^2 \phi_1 \\
  x_2 &= r \sin^2 \phi_1 \cos^2 \phi_2 \\
  x_3 &= r \sin^2 \phi_1 \sin^2 \phi_2 \cos^2 \phi_3 \\
  x_{n-1} &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \cos^2 \phi_{n-1} \\
  x_n &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \sin^2 \phi_{n-1}
 \end{aligned}
$$
so that $x_1 + \cdots + x_n = r$ (yes, I just took the generalized spherical coordinates and squared the angular part). These coordinates are only defined for $x_1, \cdots, x_n \geq 0$, so the angles only move in range $\left\langle 0, \pi/2 \right)$ each and $r > 0$. It's okay, as I'm only interested in these coordinates when they're positive. I want to prove that the Jacobian determinant of this transformation is (for a general $n$)
$$
J (r, \phi_1, \cdots, \phi_n) = 2^{n-1} r^{n-1} \cos \phi_1 \cdots \cos \phi_{n-1} \sin \phi_1^{2 n - 3} \sin \phi_2^{2n-5} \cdots \sin \phi_{n-1}
$$
so for some small $n$
$$
\begin{aligned}
J (r, \phi_1) &= 2 r \cos \phi_1 \sin \phi_1 \\
J (r, \phi_1, \phi_2) &= 4 r \cos \phi_1 \cos \phi_2 \sin^3 \phi_1 \sin \phi_2 \\
J (r, \phi_1, \phi_2, \phi_3) &= 8 r \cos \phi_1 \cos \phi_2 \cos \phi_3 \sin^5 \phi_1 \sin^3 \phi_2 \sin \phi_3 \\
\end{aligned}
$$
What's my best shot at proving this formula for general $n$? I only spotted this pattern while playing with Mathematica, so I'm not even sure it's correct, but I hope it is.
I wanted to try induction, but adding one more coordinate $x_{n+1}$ actually changes the form for $x_n$ ($x_n$ will end with $\cdots \sin^2 \phi_{n-1} \cos^2 \phi_n$ instead of just $\sin^2 \phi_{n-1}$), so it does not help that much. I wonder if it's useful to look at the metric tensor first, rather than the Jacobian determinant itself, as Jacobian determinant = det $g$ anyway. I didn't spot any useful pattern in the metric tensor itself for small $n$, as it has all off-diagonal elements in the new coordinates (because the new coordinates are not orthogonal).
 A: The proof can be done inductively using the volume forms. Let $\omega_n$ be
$$
\omega_n = \mathrm{d} x_1 \wedge \cdots \wedge \mathrm{d} x_n = J_n (r, \phi_1, \cdots, \phi_{n-1}) \, \mathrm{d} r \wedge \mathrm{d} \phi_1 \wedge \cdots \wedge \mathrm{d} \phi_{n-1}
$$
Now let's add another coordinate so we have $n+1$ coordinates, in the following form
$$
 \begin{aligned}
  x_1 &= r \cos^2 \phi_1 \\
  x_2 &= r \sin^2 \phi_1 \cos^2 \phi_2 \\
  x_3 &= r \sin^2 \phi_1 \sin^2 \phi_2 \cos^2 \phi_3 \\
  &\;\;\vdots \\
  x_{n-1} &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \cos^2 \phi_{n-1} \\
  x_n &= r \sin^2 \phi_1 \cdots \sin^2 \phi_{n-2} \sin^2 \phi_{n-1}
 \end{aligned} \quad \to \quad
 \begin{aligned}
  x_1^\prime &= x_1 \\
  x_2^\prime &= x_2 \\
  x_3^\prime &= x_3 \\
  &\;\;\vdots \\
  x_{n-1}^\prime &= x_{n-1} \\
  x_n^\prime &= x_n \cos^2 \phi_n \\
  x_{n+1}^\prime &= x_n \sin^2 \phi_n \\
 \end{aligned}
$$
Now let's calculate the form $\omega_{n+1}$
$$
\omega_{n+1} = \mathrm{d} x^\prime_1 \wedge \cdots \wedge\mathrm{d} x^\prime_{n+1} = \left( \mathrm{d} x_1^\prime \wedge \cdots \wedge \mathrm{d} x_{n-1}^\prime \wedge \mathrm{d} x_n^\prime \right) \wedge \mathrm{d} x_{n+1}^\prime = \\=\left( \mathrm{d} x_1 \wedge \cdots \wedge \mathrm{d} x_{n-1} \wedge \mathrm{d} x_n \right) \wedge \left( \frac{\partial}{\partial x_n} \left( \mathrm{d} x^\prime_n \wedge \mathrm{d} x^\prime_{n+1} \right) \right) = \omega_n \wedge \left( \frac{\partial x_n^\prime}{\partial x_n} \mathrm{d} x^\prime_{n+1} - \frac{\partial x_{n+1}^\prime}{\partial x_n} \mathrm{d} x^\prime_n \right)
$$
The one-form at the end of this expression is a matter of simple calculation
$$
 \mathrm{d} x_n^\prime = \mathrm{d} x_n \cos^2 \phi_n - 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n = \cos \phi_n \left( \mathrm{d} x_n \cos \phi_n - 2 x_n \sin \phi_n \, \mathrm{d} \phi_n \right)
$$
$$
 \mathrm{d} x_{n+1}^\prime = \mathrm{d} x_n \sin^2 \phi_n + 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n = \sin \phi_n \left( \mathrm{d} x_n \sin \phi_n + 2 x_n \cos \phi_n \, \mathrm{d} \phi_n \right)
$$
$$
\frac{\partial x_n^\prime}{\partial x_n} \mathrm{d} x^\prime_{n+1} - \frac{\partial x_{n+1}^\prime}{\partial x_n} \mathrm{d} x^\prime_n = \cos^2 \phi_n \left( \mathrm{d} x_n \sin^2 \phi_n + 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n \right) - \\-\sin^2 \phi_n \left( \mathrm{d} x_n \cos^2 \phi_n - 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n \right) = 2 x_n \sin \phi_n \cos \phi_n \, \mathrm{d} \phi_n
$$
and the factor $2 x_n \sin \phi_n \cos \phi_n$ is exactly what we find when we divide $J_{n+1}$ by $J_n$.
$$
\frac{J_{n+1}}{J_n} = \frac{2^n r^n \cos \phi_1 \cdots \cos \phi_{n-1} \cos \phi_n \sin \phi_1^{2 n - 1} \sin \phi_2^{2n-3} \cdots \sin^3 \phi_{n-1} \sin \phi_n}{2^{n-1} r^{n-1} \cos \phi_1 \cdots \cos \phi_{n-1} \sin \phi_1^{2 n - 3} \sin \phi_2^{2n-5} \cdots \sin \phi_{n-1}} = \\=2 r \cos \phi_n \sin^2 \phi_1 \sin^2 \phi_2 \cdots \sin^2 \phi_{n-1} \sin \phi_n = 2 x_n \sin \phi_n \cos \phi_n
$$
