 Solution

Choose coordinates $z$ at $p$ and $w$ at $f(p)$ vanishing at those points. The local expansion is $w\circ f\circ z^{-1}(\zeta )=a{\zeta }^v+O({\zeta }^{v+1})$, $a\ne 0$. Define the branching order $v_f(p)=v$. A change of either coordinate has nonzero linear coefficient, so it preserves the leading exponent, proving well-definedness.

Ramification means $v>1$, equivalently vanishing of the derivative in local coordinates. These zeros are isolated unless the local derivative is identically zero, which by analytic continuation would make $f$ constant. An infinite set of ramification points on a compact surface would have an accumulation point, contradicting isolation. Thus that set is finite.

A nonconstant holomorphic map of compact connected Riemann surfaces is onto, since its image is open by the local form and closed by compactness. It has a finite degree $d\ge 1$. The Riemann-Hurwitz formula isIf both genera equal $g>1$, positivity of $2g-2$ forces $d=1$ and every ramification term zero. The map is bijective and has nonzero derivative everywhere. The holomorphic inverse function theorem says that a holomorphic function with nonzero derivative has a local holomorphic inverse; these inverses patch, so $f$ is biholomorphic. For $g=1$, multiplication by two on $\mathbb{C}/\Lambda$ is a degree-four holomorphic map of an elliptic curve to itself and is not biholomorphic.