I have been looking at multiplying out the gaussian integral then converting it to spherical coordinates (or the higher order equivalent) $$\pi^{3/2}=\int_0^\infty\int_0^{2\pi}\int_0^\pi r^2e^{-r^2}\sin\varphi d\varphi d\theta dr$$
$$\pi^2=\int_0^\infty\int_0^{2\pi}\int_0^\pi\int_0^\pi r^3\sin^2\varphi\sin\psi d\psi d\varphi d\theta dr$$
$$\pi^{5/2}=\int_0^\infty\int_0^{2\pi}\int_0^\pi\int_0^\pi\int_0^\pi r^4e^{-r^2}\sin^3\varphi\sin^2\psi\sin\phi d\phi d\psi d\varphi d\theta dr$$
$$\pi^3={\int_0^\infty\int_0^{2\pi}\int_0^\pi\int_0^\pi\int_0^\pi\int_0^\pi r^5e^{-r^2}\sin^4\varphi\sin^3\phi\sin^2\psi\sin\vartheta d\vartheta d\psi d\phi d\varphi d\theta dr}$$ We can see that for an integral of $n$ dimensions we have: $$I_1(n)=\int_0^\infty r^{n-1}e^{-r^2}dr=\frac 12\int_0^\infty u^{\frac{n-2}2}e^{-u}du=\frac 12\Gamma\left(\frac n2\right)=2^{3-2n}\sqrt{\pi}\frac{(2n-3)!}{(n-2)!}$$ then there will always be the integral: $$I_2=\int_0^{2\pi}d\theta=2\pi$$ Then the remainder of the integrals will be of the form: $$I_3=\int_0^\pi\sin^\alpha(\phi)d\phi$$ where of order $n$ we will have $\alpha=1,2...(n-2)$. If we combine all this we get: $$\pi^{n/2}=2^{4-2n}\pi^{3/2}\frac{(2n-3)!}{(n-2)!}\prod_{k=1}^{n-2}\left(\int_0^\pi\sin^k\phi_k d\phi_k\right)$$ I was wondering if I can use this to find a nice product for the integrals of the powers of sine. Rearranging I got: $$\prod_{k=1}^n\left(\int_0^\pi\sin^k\phi_k d\phi_k\right)=\pi^{\frac{n-1}2}2^n\frac{n!}{(2n+1)!}$$ However I am unsure whether or not I can use Legendre's duplication formula in this way as I couldn't find the set of values for which it is valid. Any help would be great, Thanks!
I should add that I believe this is only valid for $n\ge 3$ and it clearly fails