I want to prove the relationship between the Beta function and the Gamma function, i.e.
$$ B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}. $$
I want to start from expanding $(1-x)^\beta$ using the generalized binomial theorem:
$$ \begin{aligned} B(\alpha,\beta)&=\int_0^1x^\alpha(1-x)^\beta\mathrm dx\\ &=\int_0^1x^\alpha\sum_{k\geqslant 0}(-1)^k\binom{\beta}{k}x^k\mathrm dx\\ &=\sum_{k\geqslant 0}(-1)^k\binom{\beta}{k}\frac{1}{\alpha+k+1}. \end{aligned} $$
I'm stuck here. It looks very much related to some binomial identities or high-order differences, but I just cannot find a way to proceed.
I have also tried to express it in a hypergeometric series, but I don't know how it helps.
$$ B(\alpha,\beta)=\frac{1}{\alpha+1}\sum_{k\geqslant 0}\binom{k-\beta-1}{k}\frac{(\alpha+1)^{\overline k}}{(\alpha+2)^{\overline k}}=\frac{1}{\alpha+1}\sum_{k\geqslant 0}\frac{(-\beta)^{\overline k}(\alpha+1)^{\overline k}}{1^{\overline k}(\alpha+2)^\overline k}=\frac{1}{\alpha+1}F\left(\begin{matrix}-\beta,\alpha+1\\1,\alpha+2\end{matrix}\bigg|1\right). $$