What is the expected volume of the simplex formed by n+1 points independently uniformly distributed on S^{n-1}?

What is the expected volume of the simplex formed by $n+1$ points independently uniformly distributed on ${\mathbb{S}}^{n-1}$?
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Step 1
More generally, for i points independently uniformly distributed in the interior of the n-ball and j points independently uniformly distributed on its boundary (the sphere ${\mathbb{S}}^{n-1}$), with $1\le r:=i+j-1\le n$ so that the points almost surely form an r-simplex, the moments of the volume $\mathrm{\Delta }$ of this simplex are
$E\left[{\mathrm{\Delta }}^{k}\right]=\phantom{\rule{0ex}{0ex}}\frac{1}{r{!}^{k}}{\left(\frac{n}{n+k}\right)}^{i}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\left(r+1\right)\left(n+k\right)-j+1\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\left[\left(r+1\right)n+rk\right]-j+1\right)}{\left(\frac{\mathrm{\Gamma }\left(\frac{1}{2}n\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\left[n+k\right]\right)}\right)}^{r}\prod _{l=1}^{r-1}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\left[n-r+k+l\right]\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\left[n-r+l\right]\right)}\phantom{\rule{thickmathspace}{0ex}}.$
Step 2
In our case, $i=0$, $j=n+1$, $r=n$ and $k=1$, so the desired volume is
${A}_{n}=\frac{1}{n!}\frac{\mathrm{\Gamma }\left(\frac{1}{2}{n}^{2}+\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}{n}^{2}\right)}{\left(\frac{\mathrm{\Gamma }\left(\frac{1}{2}n\right)}{\mathrm{\Gamma }\left(\frac{1}{2}n+\frac{1}{2}\right)}\right)}^{n}\prod _{l=1}^{n-1}\frac{\mathrm{\Gamma }\left(\frac{1}{2}l+\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}l\right)}\phantom{\rule{thickmathspace}{0ex}}.$
With
$\begin{array}{r}\mathrm{\Xi }\left(n\right):=\frac{\mathrm{\Gamma }\left(n+\frac{1}{2}\right)}{\mathrm{\Gamma }\left(n\right)}\end{array}$
this becomes
${A}_{n}=\frac{1}{n!}\mathrm{\Xi }\left(\frac{{n}^{2}}{2}\right)\mathrm{\Xi }{\left(\frac{n}{2}\right)}^{-n}\prod _{l=1}^{n-1}\mathrm{\Xi }\left(\frac{l}{2}\right)\phantom{\rule{thickmathspace}{0ex}}.$
Thus, with
$\begin{array}{ccccccc}n& \frac{1}{2}& 1& \frac{3}{2}& 2& \frac{9}{2}& 8\\ \mathrm{\Xi }\left(n\right)& \frac{1}{\sqrt{\pi }}& \frac{\sqrt{\pi }}{2}& \frac{2}{\sqrt{\pi }}& \frac{3\sqrt{\pi }}{4}& \frac{128}{35\sqrt{\pi }}& \frac{6435\sqrt{\pi }}{4096}\end{array}$
we find
${A}_{2}=\frac{1}{2}\frac{\mathrm{\Xi }\left(2\right)\mathrm{\Xi }\left(\frac{1}{2}\right)}{\mathrm{\Xi }\left(1\right)\mathrm{\Xi }\left(1\right)}=\frac{3}{2\pi }$
and
${A}_{3}=\frac{1}{3!}\frac{\mathrm{\Xi }\left(\frac{9}{2}\right)\mathrm{\Xi }\left(\frac{1}{2}\right)\mathrm{\Xi }\left(1\right)}{\mathrm{\Xi }\left(\frac{3}{2}\right)\mathrm{\Xi }\left(\frac{3}{2}\right)\mathrm{\Xi }\left(\frac{3}{2}\right)}=\frac{4\pi }{105}\phantom{\rule{thickmathspace}{0ex}},$
in agreement with the MathWorld values, and also
${A}_{4}=\frac{1}{4!}\frac{\mathrm{\Xi }\left(8\right)\mathrm{\Xi }\left(\frac{1}{2}\right)\mathrm{\Xi }\left(1\right)\mathrm{\Xi }\left(\frac{3}{2}\right)}{\mathrm{\Xi }\left(2\right)\mathrm{\Xi }\left(2\right)\mathrm{\Xi }\left(2\right)\mathrm{\Xi }\left(2\right)}=\frac{6435}{31104{\pi }^{2}}\phantom{\rule{thickmathspace}{0ex}}.$