$\beginend{alignat}{2 &\boldsymbol x_{a\nearrow b} &\acoloneqq x_a,\range x{a+1}b \\ &\boldsymbol x &\acoloneqq \boldsymbol x_{1\nearrow n} }$
$\displaystyle \int_{\substack{0\le \boldsymbol t \\ \land \sum \boldsymbol t \le1}} \prod_{k=1}^n t_k^{s_k-1}dt_k = \frac{\prod_{k=1}^n \Gamma(s_k)}{\Gamma(\sum \boldsymbol s +1)}$
$\beginend{align}{ I(\boldsymbol s) \acoloneqq \int_{\substack{0\le \boldsymbol t \\ \land \sum \boldsymbol t \le1}} \prod_{k=1}^n t_k^{s_k-1}dt_k \\ \\ \asupplement{-24pt}{$n=1$を開始地点に帰納法を使う。} \\ \asupplement{-24pt}{$n=1$において、} \\ I(s_1) &= \int_{0\le t_1\le1} t_1^{s_1-1} dt_1 = \frac1{s_1} = \frac{\Gamma(s_1)}{\Gamma(s_1+1)} \\ \asupplement{-24pt}{$n>1$において、} \\ I(\boldsymbol s) &= \int_{0\le t_n\le1} t_n^{s_n-1} \int_{\substack{0\le \boldsymbol t_{\nearrow n-1} \\ \land \sum \boldsymbol t_{\nearrow n-1} \le1-t_n}} \lr({\prod_{k=1}^{n-1} t_k^{s_k-1}dt_k}) dt_n \\&\quad \boldsymbol u_{\nearrow n-1} \coloneqq \frac{\boldsymbol t_{\nearrow n-1}}{1-t_n} \\&= \int_{0\le t_n\le1} t_n^{s_n-1}(1-t_n)^{\sum \boldsymbol s_{\nearrow n-1}} dt_n \int_{\substack{0\le \boldsymbol u_{\nearrow n-1} \\ \land \sum \boldsymbol u_{\nearrow n-1} \le1}} \prod_{k=1}^{n-1} u_k^{s_k-1}du_k \\&= {\rm B}\!\lr({s_n,\sum \boldsymbol s_{\nearrow n-1} +1})I(\boldsymbol s_{\nearrow n-1}) \\&= \frac{\Gamma(s_n)\Gamma(\sum \boldsymbol s_{\nearrow n-1} +1)}{\Gamma(\sum \boldsymbol s +1)}I(\boldsymbol s_{\nearrow n-1}) }$