This seems to be a favorite question everywhere, including Princeton quals. How many ways are there?

Please give a new way in each answer, and if possible give reference. I start by giving two:


Ahlfors, Complex Analysis, using Liouville's theorem.
Courant and Robbins, What is Mathematics?, using elementary topological considerations.


I won't be choosing a best answer, because that is not the point.

Here is a standard algebraic proof. It suffices to show that if $L/\mathbb{C}$ is a finite extension, then $L=\mathbb{C}$. By passing to a normal closure we assume that $L/\mathbb{R}$ is Galois with Galois group $G$. Let $H$ be the Sylow-2 subgroup of $G$ and $M=L^H$.

By the Fundamental Theorem of Galois Theory, $M/\mathbb{R}$ has odd degree. Let $\alpha\in M$ and $f(x)$ be its minimal polynomial. Then $f(x)$ has odd degree and by the Intermediate Value Theorem, a real root. As $f(x)$ is irreducible, it must have degree one. Then $\alpha\in\mathbb{R}$ and $M=\mathbb{R}$. So $G=H$ is a 2-group. Then $G_1:=Gal(L/\mathbb{C})$ is a 2-group as well. 

Assuming that $G_1$ is not trivial, there must exist a degree 2 subextension $K$ of $\mathbb{C}$. But every quadratic complex polynomial has a root (by the quadratic formula), so we have a contradiction.
