I assume you have basic knowledge of complex analysis? Since you'll need to know what it implies for a function to be analytic and as you can see, you'll be using Liouville's theorem as well.

Let's state Liouville's theorem first.

If is bounded and analytic for every , then f is a constant function.

This follows rather easy from Cauchy's inequality (one of the many ) if you write f(z) as a power series, which is possible due to its analyticity.

Now the fundamental theorem of algebra, with a bit more detail in the proof.

Proof

Letf(z) be a non-constant polynomial over C and suppose that .

This makes 1/f(z) analytic everywhere. We then have

But this implies that there exists an R > 0 for which we have (take epsilon = 1 in the definition):

if |z| > R. Becausef(z) is continuous over the disc , so is 1/f(z) and thus bounded there (since the disc is a compact set) which then implies that it is bounded on the complex plane.

But a bounded and analytic function is a constant function, according to Liouville's theorem, hence 1/f(z) is constant and so isf(z) which is a contradiction with our assumption thatf(z) was non-constant.

QED.