 Solution

A function element is a pair $(f,D)$ where $D$ is a nonempty disk and $f$ is holomorphic there; the germ at its center captures its local information. Elements continue one another when their functions agree on an overlapping connected neighborhood. A complete analytic function is a maximal collection reachable from one element by chains of analytic continuation, including all further continuations of its members.

If $f^n=p$ in the initial disk, the identity theorem propagates this identity across every overlapping disk in a continuation chain. Induction along the chain proves every element of the complete analytic function also satisfies ${\tilde{f}}^n=p$.

Put $F(z,w)=w^n-p(z)$. Its gradient is $(-p^{\prime }(z),n{w}^{n-1})$. At a point of the curve with $w\ne 0$, the second entry is nonzero; at $w=0$, the root of $p$ is simple and the first entry is nonzero. Thus the holomorphic implicit function theorem supplies charts. Its precise local statement is that a holomorphic equation with nonzero derivative in one variable can locally be solved uniquely and holomorphically for that variable as a function of the other. For $w\ne 0$, use $z$ as coordinate and write $w=w(z)$; at a simple root use $w$ as coordinate and write $z=z(w)$. The transition maps are holomorphic with nonzero derivatives, giving a Riemann surface. The two projections are holomorphic because they are holomorphic expressions in either kind of chart.

Away from the zeros of $p$, the projection onto $z$ is an unramified $n$-sheeted covering. A germ of a root corresponds to $(z,f(z))$ and its analytic continuations follow this covering. Near a simple root, $z-z_0$ is a nonzero analytic factor times $w^n$; a circuit about $z_0$ cyclically permutes all $n$ branches. For nonconstant square-free $p$, these circuits connect all sheets. Hence the germ surface $S(F)$ is the curve with the points over the roots removed, and $C$ is its branched completion obtained by adding those ramification points. An ordinary function element in the variable $z$ cannot include such a point: the zero order of a holomorphic $n$th power is divisible by $n$, whereas $p$ has zero order one. If a convention already includes these branch points in the Riemann surface of a multivalued function, that completed surface is $C$ itself. For a nonzero constant $p$, the curve has $n$ separate components, and the chosen element determines just one of them. The identically zero polynomial is excluded by the non-singularity premise.