I will sketch the proof for the Jacobian by induction on $n$, thus in a different fashion compared to What is the metric tensor on the n-sphere (hypersphere)? (which perhaps is smoother, though).
For $n=2$, we have the transformation law $A_2$
$$
x^1 = \rho\cos\phi\\
x^2 = \rho\sin\phi.
$$
Hence
$$
\frac{\partial(x^1,x^2)}{\partial(\rho,\phi)}=
\left(
\begin{array}[cc]
\ \cos\phi & -\rho \sin\phi\\
\sin\phi & \rho \cos\phi
\end{array}
\right)
$$
and the Jacobian is $J_2=\rho(\cos^2\phi+\sin^2\phi)=\rho$, i.e. $dx^1 dx^2=\rho\, d\rho d\phi$.
The idea of induction works as follows: for $n=3$, instead of using the transformation $A_{3}$, given by
$$
x^1 = \rho \cos\phi \sin\theta\\
x^2 = \rho \sin\phi \sin\theta\\
x^3 = \rho \cos\theta,
$$
directly, one uses the two combined transformations $A_{23}$, given by
$$
z^1 = \rho \cos\phi\\
z^2 = \rho \sin\phi\\
\theta=\theta,
$$
with Jacobian $J_2=\rho$, and, letting $|z|\equiv\sqrt{(z^1)^2+(z^2)^2}=\rho$, $A_{3z}$, given by
$$
x^1 = z^1 \sin\theta\\
x^2 = z^2 \sin\theta\\
x^3 = |z| \cos\theta.
$$
Now,
$$
\frac{\partial(x^1,x^2,x^3)}{\partial(z^1,z^2,\theta)}=
\left(
\begin{array}[ccc]
\ \sin\theta & 0 & z^1\cos\theta\\
0 & \sin\theta & z^2\cos\theta\\
z^1|z|^{-1}\cos\theta & z^2 |z|^{-1}\cos\theta & -|z|\sin\theta
\end{array}
\right)
$$
which gives, expanding with respect to the last line,
$$
J_{3z}=
z^1|z|^{-1}\cos\theta(z^1\cos\theta\sin\theta)+
z^2 |z|^{-1}\cos\theta(z^2\sin\theta\cos\theta)+
|z|\sin\theta \sin^2\theta\\
=|z|\sin\theta
$$
so that, being $A_3=A_{3z}\circ A_{23}$,
$$
J_3 = J_2 \cdot J_{3z} = \rho |z|\sin\theta = \rho^2 \sin\theta,
$$
i.e. $dx^1dx^2dx^3 = \rho^2\sin\theta\, d\rho d\theta d\phi$.
Let $A_n$ be
$$
x^1 = \rho \cos\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\
x^2 = \rho \sin\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\
x^3 = \rho \cos\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\
\ldots\\
x^{n-1}=\rho \cos\theta_{n-3}\sin\theta_{n-2}\\
x^n = \rho \cos\theta_{n-2};
$$
the transformation itself is built recursively: at the $n$th step, multiply the $n-1$ old coordinates by sine of the new angle and add a new coordinate $x^n$ as $\rho$ times the cosine of the new angle.
In the same spirit of the above calculation, let $A_{(n-1)n}$ be
$$
z^1 = \rho \cos\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\
z^2 = \rho \sin\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\
z^3 = \rho \cos\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\
\ldots\\
z^{n-1}=\rho \cos\theta_{n-3}\\
\theta_{n-2} = \theta_{n-2}
$$
which, by induction, will have Jacobian $J_{n-1} = \rho^{n-2}\sin\theta_{1}\sin^2\theta_{2}\ldots\sin^{n-3}\theta_{n-3}$.
Let $A_{nz}$ be
$$
x^1 = z^1 \sin\theta_{n-2}\\
x^2 = z^2 \sin\theta_{n-2}\\
\ldots\\
x^{n-1} = z^{n-1} \sin\theta_{n-2}\\
x^n = |z| \cos\theta_{n-2}.
$$
The Jacobi matrix reads
$$
\frac{\partial(x^1,x^2,\ldots,x^{n-1},x^n)}{\partial(z^1,z^2,\ldots,z^{n-1},\theta_{n-2})}
=
\left(
\begin{array}[ccccc]
\ \sin\theta_{n-2} & 0 & \ldots & 0 & z^1\cos\theta_{n-2}\\
0 & \sin\theta_{n-2} & \ldots & 0 & z^2\cos\theta_{n-2}\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & \ldots & \sin\theta_{n-2} & z^{n-1}\cos\theta_{n-2}\\
z^1|z|^{-1}\cos\theta_{n-2} & z^2|z|^{-1}\cos\theta_{n-2} & \ldots & z^{n-1}|z|^{-1} & -|z|\sin\theta_{n-2}
\end{array}
\right),
$$
where $$|z|\equiv \sqrt{\sum_{k=1}^{n-1}(z^k)^2 }=\rho.$$
Again, the Jacobian for $A_{nz}$ is easily expressed expanding with respect to the last row
$$
J_{nz} = z^1|z|^{-1}\cos\theta_{n-2} z^1\cos\theta_{n-2} \sin^{n-2}\theta_{n-2}\\
+ z^2 |z|^{-1}\cos\theta_{n-2} z^2\cos\theta_{n-2}\sin^{n-2}\theta_{n-2}\\
+ \ldots\\
+ z^{n-1}|z|^{-1}\cos\theta_{n-2} z^{n-1}\cos\theta_{n-2}\sin^{n-2}\theta_{n-2}\\
+ |z| \sin^2\theta_{n-2}\sin^{n-2}\theta_{n-2}\\
= |z| \sin^{n-2}\theta_{n-2}.
$$
Finally $A_{n} = A_{nz}\circ A_{(n-1)n}$ and hence
$$
J_n = J_{n-1}\cdot J_{nz} = \left( \rho^{n-2} \prod_{k=0}^{n-3}\sin^k\theta_{k}\right) |z| \sin^{n-2}\theta_{n-2}\\
=\rho^{n-1} \prod_{k=0}^{n-2} \sin^k\theta_k,
$$
where, for notational convenience $\theta_0\equiv \pi/2$.
This proves the Jacobian formula for any $n$.
If I recall correctly, any transformation with nonsingular Jacobi matrix gives rise to a local diffeomorphism, therefore our derivation above proves that this change of coordinates is good, except at the “north poles” and “south poles” $\theta_k =0,\pi$. To see that this indeed covers the whole $\mathbb{R}^n$, with the exception of such singularities, one can work out explicitly the inversion formulas, expressing the hyperspherical coordinates in terms of $x^1,\ldots,x^n$. This can be done, again, recursively and I think it is nicely given on https://en.wikipedia.org/wiki/N-sphere.
I hope it helps!
