Theorem: Fundamental Theorem of Algebra

Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.
That is, the field of complex numbers is algebraically closed.

Proof:
Let $f$ be a polynomial with coefficients in $\R$, and let $f$ have degree $n$.
By Fundamental Theorem of Arithmetic, there exists $k \in \N$ such that:
:$n = 2^k \cdot m$ 
with $m$ an odd number. 
We proceed by induction on $k$.
Note that $f$ has roots $\alpha_1, \ldots, \alpha_n$ in some splitting field.
It suffices to show that one of these roots is a complex number.
=== Base case ===
$k = 0$
$f$ is a polynomial with real coefficients and odd degree. Hence, $f$ has a root in $\R \subset \C$ by Rolle's Theorem. 
=== Induction hypothesis ===
For some integer $k > 0$, every polynomial with real coefficients of degree $2^{k - 1} \cdot m$, with $m$ an odd integer, has a root in $\C$.
For each nonzero $\lambda \in \Z$, construct the polynomial
:$\ds \map {g_\lambda} x = \prod_{i \mathop < j} \paren {x - \paren {\alpha_i + \alpha_j} - \lambda \alpha_i \alpha_j} $
By Viète's Formulas, each coefficient of $g_\lambda$ is a symmetric polynomial in:
:$\alpha_i + \alpha_j + \lambda \alpha_i \alpha_j$
Thus, the coefficients of $g_\lambda$ are invariant under permutations of the roots of $g_\lambda$.
But a permutation of the roots of $g_\lambda$ is equivalent to permuting the indices $\closedint 1 n$ of the roots of $f$.
Thus, the coefficients of $g_\lambda$ are invariant under permutations of the roots of $f$.
By definition, each coefficient of $g_\lambda$ is a symmetric polynomial in $\set {\alpha_1, \ldots, \alpha_n}$. 
By Fundamental Theorem of Symmetric Polynomials, each coefficient of $g_\lambda$ is a polynomial in the elementary symmetric polynomials in $\set {\alpha_1, \ldots, \alpha_n}$.
But by Viète's Formulas, these are exactly the coefficients of $f$.
Thus, each coefficient of $g_\lambda$ is a polynomial in the coefficients of $f$, which are real numbers.
Since Real Numbers form Field, the coefficients of $g_\lambda$ are real numbers.
Further, $g_\lambda$ has one linear factor for each unordered pair of roots of $f$.
Thus, the degree of $g_\lambda$ is 
{{begin-eqn}}
{{eqn | l = \binom n 2
      | r = \frac {n \cdot \paren {n - 1} } 2
      | c = Binomial Coefficient with Two
}}
{{eqn | r = \frac {2^k \cdot m \cdot \paren {2^k \cdot m - 1} } 2
      | c = expression of $n$
}}
{{eqn | r = 2^{k - 1} \cdot m \cdot \paren {2^k \cdot m - 1}
}}
{{end-eqn}}
Since $m$ and $2^k \cdot m - 1$ are odd, the degree of $g_\lambda$ is of the form:
:$2^{k-1} \cdot m'$
with:
:$m' = m \cdot \paren {2^k \cdot m - 1}$ 
an odd integer.
Combined with the fact that $g_\lambda$ has real coefficients, the induction hypothesis implies that:
:$g_\lambda$
has a root in $\C$.
Thus, we have infinitely many polynomials $g_\lambda$ with a complex root of the form:
:$\alpha_i + \alpha_j + \lambda \alpha_i \alpha_j$ 
for some $i, j\in \closedint 1 n$.
Since there are infinitely many polynomials and only finitely many pairs $\tuple {i, j}$, it follows from the Infinite Pigeonhole Principle that there exist distinct $\lambda, \mu \in \Z$ and a particular pair of roots $\alpha_x, \alpha_y$ such that:
:$\alpha_x + \alpha_y + \lambda \alpha_x \alpha_y \in \C$
and
:$\alpha_x + \alpha_y + \mu \alpha_x \alpha_y \in \C$
Since Complex Numbers form Field, the difference of these numbers:
:$\paren {\lambda - \mu} \alpha_x \alpha_y \in \C$
Since $\lambda$ and $\mu$ are distinct:
:$\lambda - \mu \ne 0$
and so:
:$\frac {\paren {\lambda - \mu} \alpha_x \alpha_y}{\lambda - \mu} = \alpha_x \alpha_y \in \C$
Since:
:$\alpha_x \alpha_y \in \C$
it follows that:
:$\lambda \alpha_x \alpha_y \in \C$
and so:
:$\alpha_x + \alpha_y + \lambda \alpha_x \alpha_y - \lambda \alpha_x \alpha_y = \alpha_x + \alpha_y \in \C$
Thus:
:$\map h x = x^2 - \paren {\alpha_x + \alpha_y} x + \alpha_x \alpha_y$ 
is a quadratic equation with complex coefficients.
By Viète's Formulas, the roots of $\map h x$ are $\alpha_x$ and $\alpha_y$.
By Solution to Quadratic Equation, the roots of $\map h x$ are:
:$\dfrac {- \alpha_x - \alpha_y \pm \sqrt{\paren {\alpha_x + \alpha_y}^2 - 4 \alpha_x \alpha_y} } 2$
which are complex numbers by Square Root of Complex Number in Cartesian Form and Complex Numbers form Field.
Thus, $\alpha_x$ and $\alpha_y$ are complex numbers.
Since $\alpha_x$ is a root of $f$, this means $f$ has a complex root. Thus, any polynomial with real coefficients has a root in $\C$.
Now let:
:$\ds \map f x = \sum_{k \mathop = 0}^n c_k x^k$ 
be a polynomial with complex coefficients, and let $\map {\overline{f} } x$ denote:
:$\ds \sum_{k \mathop = 0}^n \overline{c_k} x^k$
where an overline denotes a complex conjugate.
Observe:
{{begin-eqn}}
{{eqn | l = \overline {\map f x \map {\overline f} x}
      | r = \map {\overline f} x \overline {\map {\overline f} x}
      | c = Conjugation of Polynomials Preserves Multiplication
}}
{{eqn | r = \overline {\map f x} \map f x
}}
{{end-eqn}}
Thus:
:$\map h x = \map f x \map {\overline f} x$ 
equals its conjugate, and so has real coefficients by Complex Number equals Conjugate iff Wholly Real.
Since $\map h x$ has real coefficients, it has a complex root, $\alpha$.
By Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs:
:$\overline{\alpha}$ 
is also a root of $h$.
Since:
:$\map h \alpha = \map f \alpha \map {\overline f} \alpha = 0$
it follows that $\alpha$ is a root of either $f$ or $\overline f$.
If $\alpha$ is a root of $f$, then we are done.
If $\alpha$ is a root of $\overline f$, then:
:$\map {\overline f} \alpha = 0$
and so
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \overline {c_k} \alpha^k
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^n c_k \overline \alpha^k
      | c = taking conjugates, applying Product of Complex Conjugates
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \map f {\overline \alpha}
      | r = 0
}}
{{end-eqn}}
as desired.
{{qed}}