I'm not sure how elementary/algebraic you will consider this. Let $k$ be an algebraically closed field of characteristic $0$.
Let $W \to V$ be an etale map. Assume $W$ is connected; a general etale map will then be the disjoint union of several examples of this sort. Since $W$ is etale over $V$, we know that $W$ is smooth and one dimensional. Let $\overline{W}$ be the complete curve containing $W$, so we have a map $\overline{W} \to \mathbb{P}^1$. Let the degree of this map be $n$; let $g$ be the genus of $\overline{W}$; let $e^0_1$, ..., $e^0_r$ be the ramification degrees of the points over $0$ and let $e^{\infty}_1$, ..., $e^{\infty}_s$ be the ramification degrees of the points over $\infty$.
The Riemann-Hurwitz formula gives $2g-2 = -2n + \sum (e^0_i-1) + \sum (e^{\infty}_i-1).$ (This is the step which is invalid in positive characteristic.) The right hand side is $-2n + \sum e^0_i + \sum e^{\infty}_i - r -s = -2n+n+n-r-s=-r-s.$ So $2g+r+s=2.$
But $g \geq 0$ and $r$ and $s \geq 1$. So this can only hold if $g=0$ and $r=s=1$. The fact that $g=0$ means that $\overline{W} \cong \mathbb{P}^1_k$. The fact that $r=s=1$ means that there is one point of $\overline{W}$ lying over $0$, and one point lying over $\infty$; without loss of generality, let those points be $0$ and $\infty$.
So our map is of the form $t \mapsto p(t)/q(t)$ for some relatively prime polynomials $p$ and $q$, and the preimages of $0$ and $\infty$ are $0$ and $\infty$. So the only root of $p$ can be $0$, and $q$ can have no roots at all. We conclude that our map is of the form $t \mapsto a t^n$, as desired.
You definitely can give purely algebraic proofs that every curve embeds in a complete curve, and of Riemann-Hurwitz. I feel like one should be able to give pretty elementary ones, but I don't know a reference which does it in an elementary way.