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