The proofs of the fundamental Theorem of Algebra [closed]

There are many proofs of the fundamental theorem of algebra.

Which are the most beautiful proofs?

Let $P\in\mathbb{C}[X]$ of degree greater than $1$. Assume by contradiction that $P$ has no roots in $\mathbb{C}$ and let define: $$f:=\frac{1}{P}.$$ $f$ is holomorphic on $\mathbb{C}$ and is bounded since: $$\lim_{z\to\pm\infty}|f(z)|=0.$$ Therefore, using Liouville's theorem, $f$ is constant and so is $P$, a contradiction.

How about a proof in two parts:

$\bullet$ If $R$ is a real-closed field, then $R[X]/(X^2+1)$ is algebraically closed.

$\bullet$ $\mathbb R$ is a real-closed field.

In my opinion, the most beautiful proofs are the purely algebraic ones. We need a little analysis, but only enough to show that every odd degree polynomial over $\mathbb{R}$ has a root, which follows from the Intermediate Value Theorem (we can't really do better than this, since we must use completeness of $\mathbb{R}$ in some way for the theorem to hold).

Suppose that $\mathbb{R}\subset F$ is a finite extension of fields. If we can show that it has degree $\leq 2$, then the FTA follows.

Let $G$ be the Galois group $\operatorname{Gal}(F/\mathbb{R})$, and let $H$ be a Sylow $2$-subgroup of $G$. $H$ has odd index, so the fixed field $F^H$ of $H$ has odd degree. But a nontrivial extension of $\mathbb{R}$ of odd degree would give us an odd-degree polynomial with no root, so we have $F^H = \mathbb{R}\implies H=G$, so $G$ is a $2$-group.

If $|G|=1$ or $|G|=2$, we are done. If $|G|\geq 4$, then $G$ has a subgroup of index $2$ containing a subgroup of index $4$ (by general facts about $p$-groups). By Galois, these correspond to a degree $2$ extension of $\mathbb{C}$, which is impossible by the quadratic formula.