 Solution

Write the Papperitz symbol by listing pairs of exponents at $0,1,\infty$. For $F(a,c-b;c;w)$ they are $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. Under $w=z/(z-1)$ the points $0,1,\infty$ in the $z$-plane correspond respectively to $0,\infty ,1$ in the $w$-plane. Thus the pulled-back exponents are $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity.

Multiplication by $(1-z{)}^{-a}$ shifts the exponents at one by $-a$ and those at infinity by $+a$, yieldingThis is the original Gauss hypergeometric equation. Both functions are analytic and normalized to one at zero, so uniqueness of that normalized local solution proves the Pfaff transformationContinue the identity using the branch $|\arg (1-z)|<\pi$. As usual, the parameters must admit the normalized hypergeometric solution; excluded singular parameter values are treated only when their limiting solution exists.