Proof on representing the incomplete beta function as a hypergeometric function

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The incomplete beta function is defined as

$$B(x;c,d) = \int_0^xt^{c-1}(1-t)^{d-1}dt.$$

The hypergeometric representation of the incomplete Beta function is given by \begin{equation} B\left(x; c,d \right)=\frac{x^c}{c}{}_2F_1\left( c,1-d;1+c;x \right) \end{equation}

where \begin{equation} B(b,c-b)\,_2F_1(a,b;c;z) = \int_0^1 x^{b-1} (1-x)^{c-b-1}(1-zx)^{-a} \, dx \end{equation}

B(b,c-b) the [ beta function](https://en.wikipedia.org/wiki/Beta_function

Anyone know a nice elementary proof to this? Preferably one that doesn't expand the series out, and that someone with two semester of calculus would understand. But if not, what is the standard way to prove this?

To be clear, these two integral are equal, how can I show it