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 the proof by Pukhlikov (1997) at
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mp&paperid=6&option_lang=eng
which Ilya mentioned as being only in Russian so far. What I present below is not a literal translation (as if anyone on this site cares...).
The argument will use only real variables: there is no use of complex numbers anywhere. The goal is to show for every $n \geq 1$ that each monic polynomial of degree $n$ in ${\mathbf R}[X]$ is a product of linear and quadratic polynomials. This is clear for $n = 1$ and 2, so from now on let $n \geq 3$ and assume by induction that nonconstant polynomials of degree less than $n$ admit factorizations into a product of linear and quadratic polynomials.
First, some context: we're going to make use of proper mappings. A complex-variable proof on this page listed by Gian depends on the fact that a nonconstant one-variable complex polynomial is a proper mapping $\mathbf C \rightarrow \mathbf C$. Of course a nonconstant one-variable real polynomial is a proper mapping $\mathbf R \rightarrow \mathbf R$, but that is not the kind of proper mapping we will use. Instead, we will use the fact (to be explained below) that multiplication of real one-variable polynomials of a fixed degree is a proper mapping on spaces of polynomials. I suppose if you find yourself teaching a course where you want to give the students an interesting but not well-known application of the concept of a proper mapping, you could direct them to this argument.
Now let's get into the proof. It suffices to focus on monic polynomials and their monic factorizations. For any positive integer $d$, let $P_d$ be the space of monic polynomials of degree $d$: $$ x^d + a_{d-1}x^{d-1} + \cdots + a_1x + a_0. $$ By induction, every polynomial in $P_1, \dots, P_{n-1}$ is a product of linear and quadratic polynomials. We will show every polynomial in $P_n$ is a product of a polynomial in some $P_k$ and $P_{n-k}$ where $1 \leq k \leq n-1$ and therefore is a product of linear and quadratic polynomials.
For $n \geq 3$ and $1 \leq k \leq n-1$, define the multiplication map $$\mu_k \colon P_k \times P_{n-k} \rightarrow P_n \ \ \text{ by } \ \ \mu_k(g,h) = gh.$$ Let $Z_k$ be the image of $\mu_k$ in $P_n$ and $$Z = \bigcup_{k=1}^{n-1} Z_k.$$ These are the monic polynomials of degree $n$ which are composite. We want to show $Z = P_n$. To achieve this we will look at topological properties of $\mu_k$.
We can identify $P_d$ with ${\mathbf R}^d$ by associating to the polynomial displayed way up above the vector $(a_{d-1},\dots,a_1,a_0)$. This makes $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ a continuous mapping. The key point is that $\mu_k$ is a proper mapping: its inverse images of compact sets are compact. To explain why $\mu_k$ is proper, we will use an idea of Pushkar' to "compactify" $\mu_k$ to a mapping on projective spaces. (In the journal where Pukhlikov's paper appeared, the paper by Pushkar' with his nice idea comes immediately afterwards. Puklikov's own approach to proving $\mu_k$ is proper is more complicated and I will not be translating it!)
Let $Q_d$ be the nonzero real polynomials of degree $\leq d$ considered
up to scaling. There is a bijection
$Q_d \rightarrow {\mathbf P}^d({\mathbf R})$ associating to a class of polynomials
$[a_dx^d + \cdots + a_1x + a_0]$ in $Q_d$ the point $[a_d,\dots,a_1,a_0]$.
In this way we make $Q_d$ a compact Hausdorff space.
The monic polynomials $P_d$, of degree $d$, embed into
$Q_d$ in a natural way and are identified in ${\mathbf P}^d({\mathbf R})$
with a standard copy of ${\mathbf R}^d$.
Define $\widehat{\mu}_k \colon Q_k \times Q_{n-k} \rightarrow Q_n$ by $\widehat{\mu}_k([g],[h]) = [gh]$. This is well-defined and restricts on the embedded subsets of monic polynomials to the mapping $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$. In natural homogeneous coordinates, $\widehat{\mu}_k$ is a polynomial mapping so it is continuous. Since projective spaces are compact and Hausdorff, $\widehat{\mu}_k$ is a proper map. Then, since $\widehat{\mu}_k^{-1}(P_n) = P_k \times P_{n-k}$, restricting $\widehat{\mu}_k$ to $P_k \times P_{n-k}$ shows $\mu_k$ is proper.
Since proper mappings are closed mappings, each $Z_k$ is a closed subset of $P_n$, so $Z = Z_1 \cup \cdots \cup Z_{n-1}$ is closed in $P_n$. Topologically, $P_n \cong {\mathbf R}^n$ is connected, so if we could show $Z$ is also open in $P_n$ then we immediately get $Z = P_n$ (since $Z \not= \emptyset$), which was our goal. Alas, it will not be easy to show $Z$ is open directly, but a modification of this idea will work.
We want to show that if a polynomial $f$ is in $Z$ then all polynomials in $P_n$ that are near $f$ are also in $Z$. The inverse function theorem is a natural tool to use in this context: supposing $f = \mu_k(g,h)$, is the Jacobian determinant of $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ nonzero at $(g,h)$? If so, then $\mu_k$ has a continuous local inverse defined in a neighborhood of $f$.
To analyze $\mu_k$ near $(g,h)$, we write all (nearby) points in $P_k \times P_{n-k}$ as $(g+u,h+v)$ where $\deg u \leq k-1$ and $\deg v \leq n-k-1$ (allowing $u = 0$ or $v = 0$ too). Then $$ \mu_k(g+u,h+v) = (g+u)(h+v) = gh + gv + hu + uv = f + (gv + hu) + uv. $$ As functions of the coefficients of $u$ and $v$, the coefficients of $gv + hu$ are all linear and the coefficients of $uv$ are all higher degree polynomials.
If $g$ and $h$ are relatively prime then every polynomial of degree less than $n$ is uniquely of the form $gv + hu$ where $\deg u < \deg g$ or $u = 0$ and $\deg v < \deg h$ or $v = 0$, while if $g$ and $h$ are not relatively prime then we can write $gv + hu = 0$ for some nonzero polynomials $u$ and $v$ where $\deg u < \deg g$ and $\deg v < \deg h$. Therefore the Jacobian of $\mu_k$ at $(g,h)$ is invertible if $g$ and $h$ are relatively prime and not otherwise.
We conclude that if $f \in Z$ can be written somehow as a product of nonconstant relatively prime polynomials then a neighborhood of $f$ in $P_n$ is inside $Z$. Every $f \in Z$ is a product of linear and quadratic polynomials, so $f$ can't be written as a product of nonconstant relatively prime polynomials precisely when it is a power of a linear or quadratic polynomial. Let $Y$ be all these "degenerate" polynomials in $P_n$: all $(x+a)^n$ for real $a$ if $n$ is odd and all $(x^2+bx+c)^{n/2}$ for real $b$ and $c$ if $n$ is even. (Note when $n$ is even that $(x+a)^n = (x^2 + 2ax + a^2)^{n/2}$.) We have shown $Z - Y$ is open in $P_n$. This is weaker than our hope of showing $Z$ is open in $P_n$. But we're in good shape, as long as we change our focus from $P_n$ to $P_n - Y$. If $n = 2$ then $Y = P_2$ and $P_2 - Y$ is empty. For the first time we will use the fact that $n \geq 3$.
Identifying $P_n$ with ${\mathbf R}^n$ using polynomial coefficients, $Y$ is either an algebraic curve ($n$ odd) or algebraic surface ($n$ even) sitting in ${\mathbf R}^n$. For $n \geq 3$, the complement of an algebraic curve or algebraic surface in ${\mathbf R}^n$ for $n \geq 3$ is path connected, and thus connected.
The set $Z-Y$ is nonempty since $(x-1)(x-2)\cdots(x-n)$ is in it. Since $Z$ is closed in $P_n$, $Z \cap (P_n - Y) = Z - Y$ is closed in $P_n - Y$. The inverse function theorem tells us that $Z - Y$ is open in $P_n$, so it is open in $P_n - Y$. Therefore $Z - Y$ is a nonempty open and closed subset of $P_n - Y$. Since $P_n - Y$ is connected and $Z - Y$ is not empty, $Z - Y = P_n - Y$. Since $Y \subset Z$, we get $Z = P_n$ and this completes Pukhlikov's "real" proof of the Fundamental Theorem of Algebra.
Mы доказывали, доказывали и наконец доказали. Ура! :)
Re Gauss's first proof, Smale pointed out the following (discussed by Weber in the collected papers of Gauss): In order to show that the zero sets of the real and imaginary parts of a complex polynomial intersect, Gauss states (paraphrasing): "If a [polynomial] curve C in the plane enters a region, it must leave it. No one to whom I have explained the meaning of this result doubts it. I will give a proof in a later paper." This seems to mean, for example, that no point has an open neighborhood in the plane meeting C in a half open interval, or in a set homeomorphic to 9. In modern terms: (A) For every p in C the number of branches containing p is even. This is true-- but to use it Gauss would have to prove it without using FTA! (A) follows from a vast generalization due independently to D. Sullivan and Deligne: (A') Let p be a point in an analytic variety X over C or R, and S the boundary of a sufficiently small ball centered at p. Then the Euler characteristic of the intersection of S and X is 0 in the complex case, and even in the real case. So Gauss's celebrated proof has an enormous gap.
There's a linear algebra proof by Harm Derksen: https://www.jstor.org/stable/3647746. You can also find the article posted here: https://math.berkeley.edu/~ribet/110/f03/derksen.pdf
Two very short proofs, mostly topological, that a nonconstant polynomial map $f:{\bf C} \to \bf C$ is surjective (joint work with Robert Palais):
(1) Complex analysis shows that $f$ is an open map (images of open sets are open). A standard estimate, $|f(z)|\to\infty$ as $|z|\to\infty$, implies$f$ is also a closed map (images of closed sets are closed). Thus $f(\bf C)$ is an open, closed, nonempty subset of the connected space $\bf C$, therefore $ f(\bf C)=\bf C$.
The openness of $f$ is nontrivial, but it can be replaced by elementary algebra and topology:
(2) The set $K$ of roots of $f'$ is finite. The inverse function theorem shows that the set $A:=f({\bf C})\setminus f(K)$ is open, with finite boundary $A'=f(K) \setminus A$ because $f$ is closed. Thus $A$ has closure $\bar A = f({\bf C})$. Since a finite set cannot disconnect the plane, $\bar A = \bf C$.
A nice feature of these proofs is that they have straightforward (but not trivial) generalizations to higher dimensions:
Theorem: Every nonconstant, closed, holomorphic map between connected, complex n-dimensional manifolds, is surjective.
This is one of the proofs currently on the nLab. My goal in writing it was to see how elementary I could make it, that if you squint a little it might have been a proof from the late $18^{th}$ century. I think it's pretty close; it uses the Bolzano-Weierstrass theorem, proven by Bolzano in 1817. (It's also similar in form to the answer by Timothy Gowers https://mathoverflow.net/a/16671/2926, but with more detail, suitable perhaps for a presentation to an advanced calculus class.)
Let $f\colon \mathbb{C} \to \mathbb{C}$ be a nonconstant polynomial mapping, and suppose $f$ has no zero.
Let $s$ be the infimum of values ${|f(z)|}$; choose a sequence $z_1, z_2, z_3, \ldots$ such that ${|f(z_n)|} \to s$. Since $\lim_{z \to \infty} f(z) = \infty$, the sequence $z_n$ must be bounded; by the Bolzano-Weierstrass theorem it has a subsequence $z_{n_k}$ that converges to some point $z_0$. Then ${|f(z_{n_k})|}$ converges to ${|f(z_0)|}$ by continuity, and converges to $s$ as well, so ${|f(z)|}$ attains its absolute minimum $s$ at $z = z_0$. By supposition, $f(z_0) \neq 0$.
The polynomial $f$ may be uniquely written in the form $$f(z) = f(z_0) + g(z)(z - z_0)^n$$ where $g$ is polynomial and $g(z_0) \neq 0$. Put $$F(z) = f(z_0) + g(z_0)(z - z_0)^n$$ and choose $\delta \gt 0$ small so that $${|z - z_0|} = \delta \Rightarrow {|g(z) - g(z_0)|} \lt {|g(z_0)|}.$$
$F$ maps the circle $C = \{z : {|z - z_0|} = \delta\}$ onto a circle of radius $r = {|g(z_0)|}\delta^n$ centered at $f(z_0)$. (This uses the fact that any complex number has an $n^{th}$ root; an $18^{th}$ century mathematician might have invoked Euler's formula or De Moivre's formula.) Choose $z' \in C$ so that $F(z')$ is on the line segment between the origin and $f(z_0)$ (we can always choose $\delta$ so that also $r \lt {|f(z_0)|}$). Then $${|F(z')|} = {|f(z_0)|} - r.$$ We also have $${|f(z') - F(z')|} = {|g(z') - g(z_0)|} {|z' - z_0|^n} \lt {|g(z_0)|} \delta^n = r$$ according to how we chose $\delta$ in 2. We conclude by observing the strict inequality $${|f(z')|} \leq {|F(z')|} + {|f(z') - F(z')|} \lt {|f(z_0)|} - r + r = {|f(z_0)|},$$ which contradicts the fact that ${|f(z)|}$ attains an absolute minimum at $z = z_0$.
In this post the OP wonders how one can prove the FTA using the hairy ball theorem, after having read a book of Ian Stewart containing an allusion to such a proof. The answer I gave could be added to the current list of proofs as well. There is no doubt that this proof is known. However, it does not seem to be mentioned in the other items of this list.
The main idea of proof is already present when one wants to show that a polynomial $f\in\mathbf C[z]$ of degree $2$ has a zero using the hairy ball theorem. Suppose that it does not have any zeros. Then the holomorphic vector field $$ f\tfrac{\partial}{\partial z} $$ is a nonvanishing vector field on the Riemann sphere $S^2$. This is because $f$ has a pole of order $2$ at infinity, while $\partial/\partial z$ has a zero of order $2$ at infinity. The contradiction comes from the hairy ball theorem that states that such vector fields cannot exist. The general case of a polynomial $f$ of even degree $2d$ can be proved by arguing that the vector field $$ \sqrt[d]{f\cdot\left(\tfrac{\partial}{\partial z}\right)^{\otimes d}} $$ is well defined, and nonvanishing on $S^2$ if $f$ does not have any zeros in $\mathbf C$. See here for more details.
Gauss's first proof contained an enormous gap, since he presumed facts equivalent to the Jordan curve theorem to be true. Jordan curve theorem was proven a century later.
There is a modification on Gauss's first proof that uses only basic real analysis concepts (continuity and least upper bound principle for real numbers) on the real and complex parts of a complex polynomial (which are bivariate polynomials in either $(r,\theta)$ or $(x,y)$: On Gauss's first proof of the fundamental theorem of algebra
Sorry for the self promotion, I am an author in the proof.
There is an alternative proof for FTA using "Fredholm operators on Hilbert spaces":
Assume that $P(z)=z^n+a_{n-1}z^{n-1}+\ldots+a_1 z+a_0$ has no root in $\mathbb{C}$. Then for every $\epsilon$ the polynomial $Q(z)=\epsilon^nP(z/\epsilon)=z^n+\epsilon a_{n-1} z^{n-1} +\ldots+\epsilon^{n-1}a_1z+\epsilon^n a_0$ has no root in $\mathbb{C}$, too.
The space of entire Holomorphic functions $Hol(\mathbb{C)}$ is densely embedded in $\ell^2$ via $f(z)=\sum_{n=0}^{\infty} a_nz^n \mapsto (a_0,a_1,\ldots)$. We substitute "$z$" in $Q(z)$ by the shift operator on $\ell^2$. Then it turns out that every polynomial $Q(z) $ defines a bounded linear operator $Q$ on $\ell^2$ which restricts to "multiplicative operator" by polynomial $Q(z)$ on $Hol(\mathbb{C})$ so the operator $Q$ keeps $Hol(\mathbb{C})$ invariant. Moreover a non vanishing polynomial $Q(z)$ determines an operator on $\ell^2$ which restricts to a surjective operator on $Hol(\mathbb{C})$.
Note that the operator $Q\in B(\ell^2)$ described above is a perturbation of the $n$-shift operator so $Q$ is a fredholm operator of index $-n$. On the other hand it is a perturbation of the $n$-shift operator which is an isometry. So $Q$ satisfies $|Q(v)|>k|v|$ for all $v \in \ell^2$ and for some constant $k$. Then $Q$ is an injective operator so it can not be a surjective operator on $\ell^2$ since its index( $-n$ )is nonzero. On the other hand, since the restriction of $Q$ to a dense subspace of $\ell^2$ is surjective and $Q$ satisfies $|Q(v)>k|v|$, then $Q$ must be a surjective operator on $\ell^2$, a contradiction.
Added: This proof is based on the following note in arxiv however this arxived paper was not written well.(Full of typos, grammar problem, mistakes in english writing).
There seems to be a new elementary proof using only Bolzano–Weierstraß and an inequality:
I came across this article when I was pondering teaching the FTA to my multi-variable calculus class. In the end, I didn't have time to include it. It is nice in that it relies only on Green's theorem which we get through in the first semester. On the other hand, it is clear that they are largely being clever at avoiding introducing new definitions or standard theorems (for instance they use Cauchy-Riemann equations for polynomials without really saying so). I think it's nice nevertheless.
This proof uses the open mapping theorem, which I found in Narasimhan's book.
Theorem: Let $f\in C^\omega(\mathbb{C})$, and suppose that $|f(z)|\to\infty$ as $|z|\to\infty$. Then $f(\mathbb{C})=\mathbb{C}$.
Proof: Obviously $f$ is not constant, and so the open mapping theorem implies that $f(\mathbb{C})$ is open. Let us show that $f(\mathbb{C})$ is also closed. Suppose that $f(z_k)\to w\in\mathbb{C}$ as $k\to\infty$. Then $\{z_k\}$ is bounded, so taking a subsequence if necessary, there is $z\in\mathbb{C}$ such that $z_k\to z$. By continuity $f(z_k)\to f(z)$, concluding that $w=f(z)$.
Another probabilistic proof:
Pascu, Mihai N. A probabilistic proof of the fundamental theorem of algebra. (English summary) Proc. Amer. Math. Soc. 133 (2005), no. 6, 1707–1711 (electronic).
Summary: "We use Lévy's theorem on invariance of planar Brownian motion under conformal maps and the support theorem for Brownian motion to show that the range of a non-constant polynomial of a complex variable consists of the whole complex plane. In particular, we obtain a probabilistic proof of the fundamental theorem of algebra.''
It is different from the probabilistic proof already listed among answers to this question, which uses martingale convergence theorem.
I don't have it on hand, but Ronald Solomon's Abstract Algebra has an interesting proof using symmetric polynomials and induction on the 2-adic valuation of the degree.
Not quite an answer, but relevant:
Eilenberg and Niven proved that every "polynomial" in the quaternions has a root (provided it has only one term of highest degree). The trick is familiar: they show that such a polynomial is homotopic to $q\mapsto q^n$, which induces a map of degree $n$ on the one-point compactification of $\mathbb{H}$, namely $S^4$.
At the risk of being highly downvoted, I can't resist reposting my comment to Andrew L's answer (or rather, question) below:
is there a purely algebraic proof that for any non constant $P$ in $\mathbb{Q}[i][X]$ and $\epsilon>0$ in $\mathbb{Q}$, there is $q$ in $\mathbb{Q}[i]$ s.t. $|P(q)|<\epsilon$?
I think the statement above is purely algebraic, but I have to admit I'm a bit uncertain as to where the boundary between algebra and analysis falls.
Maybe I should have posted this as a comment to Gian Maria Dall'Ara very nice proof, because is a mere variation.
He uses the lemma : Any open proper map to a locally compact space surjects the connected components it reaches. Now any non-constant polynomial corestricted to its regular value locus is as in the lemma, so it is surjective.
Here is a "constructive proof" of the D'Alembert-Gauss theorem. Fix a degree $n > 0.$
Consider the $M_n$ the affine space of monic polynomials of degree $n$ and the proper map $\mathbb C^n \to M_n$, mapping a tuple $(z_1, \ldots , z_n)$ to $\Pi (X-z_i).$ Its critical values locus is non-disconnecting because it is a complex (singular) hypersurface : it is a polynomial image of the arrangement of hyperplanes { $z_i = z_j$}.
The map is therefore surjective.
Let $p$ be a complex polynomial of degree $n$.
By analytic continuation arguments applied to any branch of $p^{-1}$, it can be shown that any level curve $\Lambda=\{z:|p(z)|=\epsilon\}$ of $p$ which does not contain a zero of $p'$ is a Jordan curve.
Since $p'$ has at most $n-1$ zeros, we can find some point $z$ such that the level curve $\Lambda_z$ of $p$ which contains $z$ does not contain a critical point.
Let $D$ denote the bounded face of $\Lambda_z$. By the maximum modulus theorem applied to $f$ on $D$, $f$ has a zero in $D$. $\Box$
In elementary proofs like the one elucidated in Kevin McGerthy's answer, there is usually a step where for a given $n\in \mathbb{N}_0$ and given $\alpha\in \mathbb{C}_0$ a $z\in \mathbb{C}$ has to be found so that $Re(\alpha z^n)<0$. People usually browse very quickly over this part and implicitly use something like Euler's or De Moivre's formula or use a topological method. This step can however be accomplished in a more 'direct' way. Indeed, for this task it suffices to show that $\forall n \in \mathbb{N}_0:\,\forall \beta \in \{1,i,-1,-i\}$ the polynomial $z^n-\beta$ has a root. This can be shown using the Cartesian formula for the square root $$(x+yi)^{\frac{1}{2}}=\left(\frac{x+(x^2+y^2)^{\frac{1}{2}}}{2}\right)^{\frac{1}{2}}+\text{sgn}(y)\left(\frac{-x+(x^2+y^2)^{\frac{1}{2}}}{2}\right)^{\frac{1}{2}}i$$ and an induction argument (on $n$). As I suspect that many people find such a remark 'pedantic', I'll waste no more words on it :)
There is a proof using clutching functions over the sphere and the first Chern class. It is quite similar to the fundamental group proof of FTA. The trick is a polynomial without zeroes allows one to construct an isomorphism between a vector bundle with first Chern class $\deg d$ and a vector bundle with first Chern class $0$ using the polynomial restricted to circles concentric with the origin as clutching functions.
The details: From a polynomial $p: \mathbb{C} \to \mathbb{C}$, $p(z) = \sum_{j=0}^d a_j z^j$ we can construct a continuous family $p_t: \mathbb{C} \times[0,1] \to \mathbb{C}$ of polynomials such that $p_1(z)(z) = a_d z^d$ and $p_0(z) = a_0$. If $p$ has no zeroes, one can construct $p_t$ in such a way that $p_t$ has no zeroes on the unit circle.
This means that for a fixed $t \in [0,1]$ we can use $p_t$ restricted to the circle as a clutching function for $S^2$. Since $p_t$ is continuous family, this gives a vector bundle $E$ over $S^2 \times [0,1]$. It is a standard fact in the theory of vector bundles that $E$ restricted to $S^2 \times \{0\}$ isomorphic to $S^2 \times\{0\}$. But our construction allows us to read off that the former has first chern class $d$, while the latter has first chern class $0$. Hence $d = 0$ and $p$ must be constant.
In the recent paper
Anton, R.; Mihalache, N.; Vigneron, F., A Short ODE Proof of the Fundamental Theorem of Algebra., Math. Intelligencer (2023).
the authors give a proof based on the analysis of the attractive points of the initial value problem:
$$\varphi_z(0)=z, \quad \varphi_z^{\prime}(t)=-\frac{P\left(\varphi_z(t)\right)}{P^{\prime}\left(\varphi_z(t)\right)},$$
which is a continuous-time version of Newton's method for computing the zeros of the polynomial $P$.
Here is a proof that I think deserves to be recorded here somewhere. Of the proofs already listed it is closest to Pushkar's proof, Lucas Culler's proof, and Gian Maria Dall'Ara's highest-upvoted proof (arguably it is "just" a more complicated version of this third proof); I like the clear intuition it offers for why FTA is true, and it constructs a different interesting auxiliary space than the spaces constructed in the other proofs. Daniel Litt wrote a version of it down in 2011, and it is related to Arnold's proof of the unsolvability of the quintic which dates to 1963.
The idea is to argue that
the space of polynomials of degree $n$ with distinct roots (over a splitting field) is connected, and
the number of roots (over $\mathbb{C}$) of a polynomial with distinct roots is a locally constant function on this space.
Since e.g. $p(z) = z^n - 1$ has $n$ distinct roots over $\mathbb{C}$, it follows that every polynomial of degree $n$ with distinct roots has $n$ distinct roots over $\mathbb{C}$. This establishes the FTA in general, because if $f(z)$ is an arbitrary polynomial we can apply this result to its squarefree part $\frac{f(z)}{\gcd(f(z), f'(z))}$.
In more detail, the space $p(z) = z^n + a_{n-1} z^{n-1} + \dots a_0$ of monic polynomials of degree $n$ is $\mathbb{C}^n$. The discriminant $\Delta$ (the resultant $\text{res}(p, p')$) is a polynomial on $\mathbb{C}^n$ which is zero on a polynomial $p(z)$ iff it has repeated roots (over a splitting field). The complement $\mathbb{C}^n \setminus \{ \Delta = 0 \}$ is connected (this is the part of the argument that fails over $\mathbb{R}$), either abstractly because it has codimension $2$ or concretely by an argument involving complex lines.
Now it suffices to argue that the number of roots of a polynomial is a locally constant function on this complement. Here I am actually not sure what the cleanest way to finish the argument is. This can be done using the inverse function theorem; a calculation of a suitable Jacobian shows that in fact, on $\mathbb{C}^n \setminus \{ \Delta = 0 \}$ the roots of a polynomial are $C^{\infty}$ functions of its coefficients, so we can just follow the (distinct) roots around as we take a smooth path in $\mathbb{C}^n \setminus \{ \Delta = 0 \}$.
The argument above implies that $\mathbb{C}^n \setminus \{ \Delta = 0 \}$ can be identified with the unordered configuration space $\text{Conf}_n(\mathbb{C})$ of $n$ distinct points in $\mathbb{C}$. This is famously a classifying space for the braid group $B_n$. Essentially what we have to show to prove FTA is that the map sending a polynomial to its roots is a local system on this space. Its monodromy is given by the natural quotient map $B_n \to S_n$, and in particular is surjective; that is, moving around in a loop in $\mathbb{C}^n \setminus \{ \Delta = 0 \}$ can produce any permutation of a (separable) polynomial's roots. This is the starting point of Arnold's topological proof of the unsolvability of the quintic.
One can, as with several other proofs, see clearly where this argument fails over $\mathbb{R}$: the complement $\mathbb{R}^n \setminus \{ \Delta = 0 \}$ is no longer connected, because now $\{ \Delta = 0 \}$ only has codimension $1$ rather than codimension $2$. Its connected components are labeled by the number of real roots (equivalently, the number of complex roots); we can visualize moving around in this space by visualizing the roots, which are either real and so confined to the real line or which exist in complex conjugate pairs. At the discriminant locus $\Delta = 0$ two real roots can collide and split into a conjugate pair of complex roots, or two conjugate complex roots can collide on $\mathbb{R}$ and split into two real roots.
Gian Maria Dall'Ara's proof is a simplification of this proof where instead of varying all the coefficients we only vary the constant term. But if we do the proof this way we don't get to talk about braid group monodromy!
I'm surprised that no-one's mentioned the proof using Roueche's theorem:
Given $f,g$ holomorphic and $C$ a closed contour if $|g(z)|< |f(z)|$ on $C$ then $f$ and $f+g$ have the same number of zeros (counting multiplicity) in the interior of $C$. There's an easy proof of this using the Cauchy integral formula.
If Let $g(z) = a_{n-1} z^{n-1} + \cdots + a_0$, and $f(z) = z^n$. If $R$ is sufficiently big then $|g(z)|<|f(z)|$ on the circle of radius $R$ with the center at 0. Thus $p(z) := z^n + g(z)$ has $n$ zeros inside that circle.
[As a side note, when I was taught this by Lipman Bers, he picturesquely referred to it as the "dog on the leash theorem" -- it's essentially a winding number argument]