TITLE: Ways to prove the fundamental theorem of algebra
QUESTION [202 upvotes]: 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.

REPLY [11 votes]: Here is a variant of d'Alembert's argument using the minimum of $|p(z)|$. It has the advantage that it proves more generally the Gelfand-Mazur theorem (usually proved by complex analysis): Any Banach field $K$ over $\mathbb C$ is $\mathbb C$ itself. Indeed, this gives the fundamental theorem of algebra by applying it to any putative finite extension of $\mathbb C$.
Let $x\in K$. The map
$$
\mathbb C\to \mathbb R_{\geq 0}: z\mapsto |x-z|
$$
is continuous and gets large for $|z|$ large, so attains its minimum at some $z\in \mathbb C$. Replacing $x$ by $x-z$, we get $(\ast)$: for all $w\in \mathbb C$, one has $|x-w|\geq |x|$. We claim that then $x=0$. If not, we can rescale $x$ by a real number so that $|x|=2$. Now take some large $n$ and consider the identity
$$ \prod_{i=0}^{n-1} (x-\zeta_n^i) = x^n-1.$$
Each factor on the left is of norm at least $|x|=2$ (by $(\ast)$), so the left-hand side is of norm at least $|x-1|2^{n-1}\geq 2^n$. The right-hand side is bounded in norm by $2^n+1$. Taking the limit as $n\to \infty$, we see that $|x-1|=2$. Applying this argument to $x-1$ in place of $x$, this gives the absurd $2=|x|=|x-1|=|x-2|=\ldots=|x-5|\geq 5-|x|=3$.
Edit: This proof is due to Ostrowski, see Section 7 in "Über einige Lösungen der Funktionalgleichung $\varphi(x) \varphi(y) = \varphi(xy)$", 1916. (h/t Mohan Ramachandran!)
The proof uses the existence of infinitely many roots of unity; but $\zeta_4=i$ and if $\zeta_{2^n} = a_n + b_n i$ then $\zeta_{2^{n+1}} = a_{n+1} + b_{n+1} i$ where $a_{n+1}=\sqrt{\frac{1+a_n}2}$ and $b_{n+1}=\sqrt{\frac{1-a_n}2}$. (But one can also rewrite the proof so as to only use the existence of $\zeta_8=\frac{1+i}{\sqrt{2}}$.)