The meromorphic functions on an open disc $\Delta$ in $\mathbf{C}$ form a field $M(\Delta)$.
How to show in as elementary a way as possible that for every polynomial $P(X) = X^n + a_1 X^{n-1} +\dots + a_n$ with coefficients in $M(\Delta)$ there is an open disc $\Delta_1\subset \Delta$ and a function $f\in M(\Delta_1)$ satisfying $$ (f(z))^n + a_1(z) (f(z))^{n-1} +\dots + a_n(z) = 0 $$ at all the points $z\in \Delta_1$, where $f$ and the $a_i$ are holomorphic?
The most simple proof I'm aware of uses both the argument principle and the residue calculus, which are quite heavy an artillery already.
Replacing $P$ by $P/\gcd(P,P')$ we can assume $P$ is squarefree.
Let $h=Discriminant(P)\in M(\Delta)$, if the $a_j$ are analytic at $c$ and $h$ is non-zero and analytic at $c$ then it is non-zero analytic on a disc $\Delta_1\ni c$, for any root $P_c(b)=0$ there is a unique $f$ continuous on $\Delta_1$ such that $f(c)=b$ and $P(f)=0$ (the roots move continuously with the coefficients and the non-zero discriminant says those root curves don't intersect). Then $f$ is holomorphic thus analytic on $\Delta_1$.