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.
Let f(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. Because f(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 is f(z) which is a contradiction with our assumption that f(z) was non-constant.