Hypergeometric Function simple identity I must prove this property but I really have no idea of how to prove it:
$${}_2F_1(a,b;c;z)=(1-z)^{-a}{}_2F_1(a,c-b;c,\frac{-z}{1-z}) $$
It seems it's a 'simple' property, but I haven't been able to prove it, I tried doing this:
$$(1-z)^{-a}{}_2F_1(a,c-b;c,\frac{-z}{1-z})=\sum_{n=0}^\infty (1-z)^{-a}\frac{(a)_n(c-b)_n}{(c)_n n!}(-1)^nz^n(1-z)^{-n}$$
$$= \sum_{n=0}^\infty (1-z)^{-a}\frac{(a)_nz^n}{(c)_n n!}\sum_{k=0}^\infty(-1)^n(c-b)_n\frac{(n)_kz^k}{k!} $$
Since we want that to be equal to the original hypergeometric function, we need to cancel out the term $(1-z)^{-a}$, and also we need to get the term $(b)_n$ in the numerator, so I thought we should have: 
$$\sum_{k=0}^\infty (-1)^n(c-b)_n\frac{(n)_kz^k}{k!} =(1-z)^a(b)_n $$
But that can't be since there is not way to get an $a$ from the expression on the left. 
Any idea? Thanks and regards. 
 A: The easiest way to prove the identity is to use Euler's integral representation
$$ _{2}F_{1} (a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_{0}^{1} t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a} \ dt.$$
Making the substitution $u = 1-t$,
$$\begin{align} _{2}F_{1} (a,b;c;z) &= \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_{0}^{1} (1-u)^{b-1} u^{c-b-1} (1-z+zu)^{-a} \ du \\ &= (1-z)^{-a} \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)}  \int_{0}^{1} u^{c-b-1} (1-u)^{b-1}  \left( 1+ \frac{zu}{1-z}\right)^{-a} \ du \\ &= (1-z)^{-a} \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \frac{\Gamma(c-b)\Gamma(b)}{\Gamma(c) } {}_{2}F_{1} \left(a,c-b;c; \frac{-z}{1-z} \right) \\  &=(1-z)^{-a} {}_{2}F_{1} \left(a,c-b;c; \frac{-z}{1-z} \right) . \end{align}$$ 
EDIT:
To prove it using the series representation, first replace $b$ with $c-b$ to get the equivalent identity
$$_{2}F_{1}(a,c-b;c;z) = (1-z)^{-a} {}_{2}F_{1}\left( a,b; c; \frac{-z}{1-z} \right).  $$
Then using $(a)_{n}$ to represent the rising factorial,
$$ \begin{align} (1-z)^{-a} {}_{2}F_{1}\left( a,b; c; \frac{-z}{1-z} \right) &=(1-z)^{-a} \sum_{n=0}^{\infty} \frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac{(-1)^{n}z^{n}}{(1-z)^{n}} \frac{1}{n!}  \\  &= \sum_{n=0}^{\infty} \frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac{(-1)^{n}z^{n}}{n!} \sum_{k=0}^{\infty} \binom{a+n+k-1}{k} z^{k} \tag{1} \\ &= \sum_{n=0}^{\infty} \frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac{(-1)^{n}z^{n}}{n!} \sum_{k=0}^{\infty} \frac{\Gamma(a+n+k)}{k! \ \Gamma(a+n)} z^{k} \\ &= \sum_{j=0}^{\infty} \sum_{r=0}^{j} \frac{(a)_{r}(b)_{r}}{(c)_{r}} \frac{(-1)^{r}}{r!} \frac{\Gamma(a+j)}{(j-r)! \ \Gamma(a+r)} z^{j} \tag{2} \\ &= \sum_{j=0}^{\infty} \sum_{r=0}^{j} \frac{\Gamma(a+r)}{\Gamma(a)}\frac{(b)_{r}}{(c)_{r}} \frac{1}{r!} \frac{(-1)^{r}}{(j-r)!}  \frac{\Gamma(a+j)}{\Gamma(a+r)} z^{j} \\ &= \sum_{j=0}^{\infty}(a)_{j} \sum_{r=0}^{j} \frac{(b)_{r}}{(c)_{r}} \frac{1}{r!} \frac{(-1)^{r}}{(j-r)!} z^{j} \\ &= \sum_{j=0}^{\infty} \frac{(a)_{j}}{j!} \sum_{r=0}^{j} \frac{(b)_{r}}{(c)_{r}} \frac{(-j)_{r}}{r!} z^{j} \\ &= \sum_{j=0}^{\infty} \frac{(a)_{j} (c-b)_{j}}{(c)_{j}} \frac{z^{j}}{j!} \tag{3} \\ &= {}_{2}F_{1}(a,c-b;c;z). \end{align}$$
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$(1)$ binomial theorem
$(2)$ Cauchy product
$(3)$ Chu-Vandermonde identity
