Euler Hypergeometric transformation proof

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-a-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$

$$\textit{proof}$$

In the Pfaff transformations let \( z \to \frac{z}{z-1}\) , proved here 

$${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)$$

and

$${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$

By equating the two transformations

$$(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$

Now use the transformation \( b \to c-b \)

$$(1-z)^{-a} {}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$

Which can be reduced to

$$ {}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-a-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$

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