I'm trying to show this equation $$ \int\limits_0^\infty \mathrm{d}x_1 \dots \mathrm{d}x_n \left( 1 - \sum_{i=1}^n x_i \right)^k \Theta \left( 1 - \sum_{i=1}^n x_i \right) = \frac{\Gamma(k+1)}{\Gamma(n+k+1)} $$ The result tells me induction is probably the way to go. I tried fixing $n$ and fixing $k$ but for both cases I'm unable to isolate the additional term/integral.
I also tried to substitute $t_i^2 = x_i$ hoping I could use polar coordinates but because of the additional factor $\mathrm{d}x_i = 2 t_i \mathrm{d}t_i$ that doesn't seem to work either.
This is asking for a substitution of the form $u_i=\sum_{j=1}^i x_j$ replacing $x_i$. You then have to worry about the domain of integration: a moment's thought will tell you that it conveniently transforms to $0<u_1<u_2< \dotsb < u_n$, since the $x_i$ are all positive. Then the Jacobian is trivial, and the integral splits into $$ \int_0^{\infty} \Theta(1-u_n) (1-u_n)^k \left( \int_{0<u_1<\dotsb<u_{n}} du_1 \dotsm du_{n-1} \right) \, du, $$ and the interior integral is obviously $u_n^{n-1} V_{n-1}$, where $$V_{n-1} = \int_{0<u_1<\dotsb<u_{n-1}<1} du_1 \dotsm du_{n-1};$$ this can be done easily by induction, giving $1/(n-1)!$. Hence the integral becomes $$ \frac{1}{(n-1)!}\int_0^1 u^{n-1} (1-u)^k \, du = \frac{1}{(n-1)!}B(n,k+1) = \frac{\Gamma(k+1)}{\Gamma(n+k+1)}, $$ as desired.