ガウスの超幾何定理の証明!(タイトル詐欺かもしれない)

${\displaystyle \frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(b+n)}}$という式について考える$(n\in \mathbb{N})$
${\displaystyle \sum _{k=0}^{\mathrm{\infty }}(f(a+k)-f(a+k+1))=f(a)-\underset{n\to \mathrm{\infty }}{lim}f(a+n)}$を多用します。
${\displaystyle \frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(b+n)}}$
${\displaystyle =\sum _{s_1=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s_1)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b+s_1)\mathrm{\Gamma }(b+n)}-\frac{\mathrm{\Gamma }(a+s_1+1)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b+s_1+1)\mathrm{\Gamma }(b+n)}}$
$(f(s_1)=\frac{\mathrm{\Gamma }(a+s_1)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b+s_1)\mathrm{\Gamma }(b+n)}として{\displaystyle f(0)=\sum _{s_1=0}^{\mathrm{\infty }}f(s_1)-f(s_1+1)をしています。)}$
${\displaystyle =\sum _{s_1=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s_1)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b+s_1)\mathrm{\Gamma }(b+n)}(1-\frac{a+s_1}{a+b+s_1})}$
${\displaystyle =\sum _{s_1=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s_1)\mathrm{\Gamma }(b+1)}{\mathrm{\Gamma }(a+b+s_1+1)\mathrm{\Gamma }(b+n)}}$
${\displaystyle =\sum _{s_1=0}^{\mathrm{\infty }}\sum _{s_2=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s_1+s_2)\mathrm{\Gamma }(b+1)}{\mathrm{\Gamma }(a+b+s_1+s_2+1)\mathrm{\Gamma }(b+n)}-\frac{\mathrm{\Gamma }(a+s_1+s_2+1)\mathrm{\Gamma }(b+1)}{\mathrm{\Gamma }(a+b+s_1+s_2+2)\mathrm{\Gamma }(b+n)}}$
${\displaystyle =\sum _{s_1=0}^{\mathrm{\infty }}\sum _{s_2=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s_1+s_2)\mathrm{\Gamma }(b+2)}{\mathrm{\Gamma }(a+b+s_1+s_2+2)\mathrm{\Gamma }(b+n)}}$
これを繰り返すことで、
${\displaystyle =\sum _{s_1,s_2\cdots ,s_n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s_1+s_2\cdots +s_n)\mathrm{\Gamma }(b+n)}{\mathrm{\Gamma }(a+b+s_1+s_2\cdots +s_n+n)\mathrm{\Gamma }(b+n)}}$
${\displaystyle =\sum _{s_1,s_2\cdots ,s_n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s_1+s_2\cdots +s_n)}{\mathrm{\Gamma }(a+b+s_1+s_2\cdots +s_n+n)}}$
${\displaystyle =\sum _{s=0}^{\mathrm{\infty }}\sum _{s_1+s_2+\cdots +s_n=s}^{}\frac{\mathrm{\Gamma }(a+s)}{\mathrm{\Gamma }(a+b+s+n)}}$
${\displaystyle =\sum _{s=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)}{\mathrm{\Gamma }(a+b+s+n)}(\genfrac{}{}{0ex}{}{s+n-1}{n-1})(非負整数n個の数字の和でsにする方法:(\genfrac{}{}{0ex}{}{s+n-1}{n-1}))}$
${\displaystyle =\sum _{s=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(s+n)}{\mathrm{\Gamma }(a+b+s+n)\mathrm{\Gamma }(n)\mathrm{\Gamma }(s+1)}}$
よって、
${\displaystyle \sum _{s=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(a+s)\mathrm{\Gamma }(s+n)}{\mathrm{\Gamma }(a+b+s+n)\mathrm{\Gamma }(s+1)}=\frac{\mathrm{\Gamma }(n)\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(b+n)}}$

$\mathrm{\Gamma }(x+n)=\mathrm{\Gamma }(x)(x{)}_n$で書き換えると、

${\displaystyle \sum _{s=0}^{\mathrm{\infty }}\frac{(a{)}_s(n{)}_s\mathrm{\Gamma }(a)\mathrm{\Gamma }(n)}{(a+b+n{)}_{s}s!\mathrm{\Gamma }(a+b+n)}=\frac{\mathrm{\Gamma }(n)\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(b+n)}}$
定数を左辺に集めて、
${\displaystyle \sum _{s=0}^{\mathrm{\infty }}\frac{(a{)}_s(n{)}_s}{(a+b+n{)}_{s}s!}=\frac{\mathrm{\Gamma }(a+b+n)\mathrm{\Gamma }(b)}{\mathrm{\Gamma }(a+b)\mathrm{\Gamma }(b+n)}}$