= 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 $+a$, yielding
$$
P\left\{\begin{array}{ccc|c}0&1&\infty&z\\0&0&a&\\1-c&c-a-b&b&\end{array}\right\}.
$$
This 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 transformation>
$$
\boxed{F(a,b;c;z)=(1-z)^{-a}F\left(a,c-b;c;\frac z{z-1}\right).}
$$
Continue 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.
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