The transformation formula states that for any $\mathcal C^1(U,V)$-diffeomorphism $\Phi$ that $$\int_{V}f(v)\,\mathrm dv = \int_{U}f\big(\Phi(u)\big)\left\vert\det\big(\mathrm D\Phi(u)\big)\right\vert\,\mathrm du$$ for integrable function $f$. If $U = (0,\infty)\times(0,\pi)\times\cdots\times(0,\pi)\times(0,2\pi)\subseteq\mathbb R^n$, and $\Phi$ is the polar coordinate transform, i.e. $$\begin{align*} v_1 &= r\cos(\varphi_1)\\v_2 &= r\sin(\varphi_1)\cos(\varphi_2) \\\vdots\\v_{n-1} &= r\sin(\varphi_1)\sin(\varphi_2)\cdots\sin(\varphi_{n-2})\cos(\varphi_{n-1})\\v_n &= r\sin(\varphi_1)\sin(\varphi_2)\cdots\sin(\varphi_{n-2})\sin(\varphi_{n-1}), \end{align*}$$ then $$\det\big(\mathrm D\Phi(u)\big) = r^{n-1}\sin(\varphi_1)^{n-2}\sin(\varphi_2)^{n-3}\cdots\sin(\varphi_{n-2})$$ with $u = (r,\varphi_1,\varphi_2,\dots,\varphi_{n-1})$. If $\Psi$ denotes the $\sin$ and $\cos$ transformations on the rhs, the system becomes $v = r\Psi(\varphi_1,\varphi_2,\dots,\varphi_{n-1})$.
There is a result which states that there exists a unique measure $S$ on the $n$-dimensional unit sphere $\mathbb S^{n-1}$ such that $$\int_Vf(v)\,\mathrm d v = \int_{(0,\infty)}\int_{\mathbb S^{n-1}}f(rw)r^{n-1}\,\mathrm dS(w)\,\mathrm dr.$$
How can I show that $$\int_{U}f\big(\Phi(u)\big)\left\vert\det\big(\mathrm D\Phi(u)\big)\right\vert\,\mathrm du = \int_{(0,\infty)}\int_{\mathbb S^{n-1}}f(rw)r^{n-1}\,\mathrm dS(w)\,\mathrm dr?$$
The factor $r^{n-1}$ is already there on both sides. Furthermore, the outermost integral on both sides coincides by definition of $U$. However, how can I establish the equality $$\int_{(0,\pi)}\cdots\int_{(0,\pi)}\int_{(0,2\pi)}f\big(r\Psi(\varphi_1, \varphi_2,\dots,\varphi_{n-1})\big)\vert\sin(\varphi_1)^{n-2}\sin(\varphi_2)^{n-3}\cdots\sin(\varphi_{n-2})\vert\,\mathrm d\varphi_1\mathrm d\varphi_2\cdots\mathrm d\varphi_{n-1} = \int_{\mathbb S^{n-1}}f(rw)\,\mathrm dS(w).$$