Proof. We use Liouville’s theorem by contradiction. Suppose, if possible \(P(z)\) has no roots in the finite complex plane. Consider the function \begin {align*} f(z) & = \frac {1}{P(z)} = \frac {1}{z^n + a_{n-1}z^{n - 1} + \cdots + a_0}\\\\ & = \frac {1}{z^n}\Big [\frac {1}{1 + a_{n-1}/z + a_{n-2}/z^2 + \cdots + a_0/z^n}\Big ] \end {align*}

which must be an entire function. Take modulus both sides \[\left |f(z)\right | = \frac {1}{\left |z^n\right |}\,\frac {1}{\left |1 + \frac {a_{n-1}}{z} + \frac {a_{n-2}}{z^2} + \cdots + \frac {a_0}{z^n}\right |}\]

Now, for \(z\) on a circle of radius \(R,\, \left |z\right | = R\)

\(\lim \limits _{R\rightarrow \infty } \frac {1}{\left |z\right |^n} = 0\,\) and \(\,\lim \limits _{R\rightarrow \infty }\left |1 + \frac {a_{n-1}}{z} + \frac {a_{n-2}}{z^2} + \cdots + \frac {a_0}{z^n}\right |\)

Recall, \(\, \left |z_1 + z_2\right |\geq \Big (\left |z_1\right | - \left |z_2\right |\Big )\,\) hence \(\, \lim \limits _{R\rightarrow \infty }\left |f(z)\right | = 0\)

This limit asserts that for any \(a_n > 0\) there exists \(R>0\) such that \(\left |f(z)\right | <\varepsilon \) for \(\left |z\right | > R\) and so \(f(z)\) must be bounded outside \(\left |z\right | = R\), thus it is bounded everywhere in the finite \(\mathbb {C}-\)plane. By Liouville’s theorem, \(f(z)\) must be identically constant. This is a contradiction, thus our assumption that \(P(z)\) has no zeros is false. Thus if \(z_1\) is zero then \(\, P(z) = \big (z - z_1\big )Q_{n-1}(z),\,\) where \(Q_{n-1}(z)\) is a polynomial of degree \(n-1\).
The same argument shows that \(\, Q_{n-1}(z) = \big (z-z_2\big )Q_{n-2}(z)\,\) where \(z_1\) may or may not be equal to \(z_2\).
Repeat the process until no more factors are found, we have \[P(z) = (z-z_1)(z-z_2) \cdots (z-z_n)\] where the zeros may be real or complex and repeated zeros are counted according to their multiplicities. □