What is the expected volume of the simplex formed by $n+1$ points independently uniformly distributed on $\mathbb S^{n-1}$? I was surprised that I couldn’t find this question answered on this site (not for lack of trying). I need the result for answering Probability of random sphere lying inside the unit ball, so I’m posting it as a separate question and answer for future reference.

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

MathWorld has the answers $\frac3{2\pi}$ for $n=2$ under Circle Triangle Picking and $\frac{4\pi}{105}$ for $n=3$ under Sphere Tetrahedron Picking, but the general result is surprisingly hard to find.
 A: The question is answered in Isotropic Random Simplices by R. E. Miles (Advances in Applied Probability, 3(2), 353–382) in Theorem $2$ on p. $362$. 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 $\Delta$ of this simplex are
$$
E\left[\Delta^k\right]
=\\
\frac1{r!^k}\left(\frac n{n+k}\right)^i\frac{\Gamma\left(\frac12(r+1)(n+k)-j+1\right)}{\Gamma\left(\frac12[(r+1)n+rk]-j+1\right)}\left(\frac{\Gamma\left(\frac12n\right)}{\Gamma\left(\frac12[n+k]\right)}\right)^r\prod_{l=1}^{r-1}\frac{\Gamma\left(\frac12[n-r+k+l]\right)}{\Gamma\left(\frac12[n-r+l]\right)}\;.
$$
In our case, $i=0$, $j={n+1}$, $r=n$ and $k=1$, so the desired volume is
$$
A_n=\frac1{n!}\frac{\Gamma\left(\frac12n^2+\frac12\right)}{\Gamma\left(\frac12n^2\right)}\left(\frac{\Gamma\left(\frac12n\right)}{\Gamma\left(\frac12n+\frac12\right)}\right)^n\prod_{l=1}^{n-1}\frac{\Gamma\left(\frac12l+\frac12\right)}{\Gamma\left(\frac12l\right)}\;.
$$
With
\begin{eqnarray}
\Xi(n):=\frac{\Gamma\left(n+\frac12\right)}{\Gamma(n)}
\end{eqnarray}
this becomes
$$
A_n=\frac1{n!}\Xi\left(\frac{n^2}2\right)\Xi\left(\frac n2\right)^{-n}\prod_{l=1}^{n-1}\Xi\left(\frac l2\right)\;.
$$
Thus, with
\begin{array}{c|cc}
n&\frac12&1&\frac32&2&\frac92&8\\\hline
\Xi(n)&\frac1{\sqrt\pi}&\frac{\sqrt\pi}2&\frac2{\sqrt\pi}&\frac{3\sqrt\pi}4&\frac{128}{35\sqrt\pi}&\frac{6435\sqrt\pi}{4096}\\
\end{array}
we find
$$
A_2=\frac12\frac{\Xi(2)\Xi\left(\frac12\right)}{\Xi(1)\Xi(1)}=\frac3{2\pi}
$$
and
$$
A_3=\frac1{3!}\frac{\Xi\left(\frac92\right)\Xi\left(\frac12\right)\Xi(1)}{\Xi\left(\frac32\right)\Xi\left(\frac32\right)\Xi\left(\frac32\right)}=\frac{4\pi}{105}\;,
$$
in agreement with the MathWorld values, and also
$$
A_4=\frac1{4!}\frac{\Xi(8)\Xi\left(\frac12\right)\Xi(1)\Xi\left(\frac32\right)}{\Xi(2)\Xi(2)\Xi(2)\Xi(2)}=\frac{6435}{31104\pi^2}\;.
$$
