Prove that $$\int_{(0,+\infty)^{r-1}} \frac{x_1^{\gamma_1}\cdots x_{r-1}^{\gamma_{r-1}}}{ \left(1+x_1+\cdots+ x_{r-1}\right)^{\gamma_1+\cdots+\gamma_r+r}} \,dx_1\cdots dx_{r-1} =\frac{\gamma_1!\cdots \gamma_r!}{\left(r+\gamma_1+\cdots+\gamma_r-1\right)!}.$$
My attempt
Denote the integration by $I$. Note that $$ a^{-m}=\frac{1}{\Gamma(m)}\int_0^{+\infty} t^{m-1} e^{-at}\,dt. $$ It follows that \begin{align*} &\qquad I\cdot (\gamma_1+\cdots+\gamma_r+r-1)!\\ &= \int_{(0,+\infty)^{r-1}}\int_0^{+\infty} {x_1^{\gamma_1}\cdots x_{r-1}^{\gamma_{r-1}}}\,\, t^{\gamma_1+\cdots+\gamma_r+r-1} e^{-(1+x_1+\cdots+ x_{r-1}\,\,)t} \,dtdx_1\cdots dx_{r-1}\\ &= \int_0^{+\infty}\int_{(0,+\infty)^{r-1}} {x_1^{\gamma_1}\cdots x_{r-1}^{\gamma_{r-1}}}\,\, t^{\gamma_1+\cdots+\gamma_r+r-1} e^{-(1+x_1+\cdots+ x_{r-1}\,\,)t} \,dx_1\cdots dx_{r-1}dt\\ &= \int_0^{+\infty} \gamma_1!\cdots\gamma_{r-1}! t^{-(\gamma_1+1)-\cdots-(\gamma_{r-1}\,\,+1)} t^{\gamma_1+\cdots+\gamma_r+r-1} e^{-t}dt\\ &= \gamma_1!\cdots\gamma_{r-1}! \int_0^{+\infty} t^{\gamma_r} e^{-t}dt\\ &= \gamma_1!\cdots\gamma_{r}!. \end{align*} We are done.
I want to know more approaches to this problem since this method is not the usual way to compute integration for me. Any ideas would be highly appreciated!