[proofplan] We argue by contradiction. If $p$ has no root in $\mathbb{C}$, then $1/p$ is an entire [function](/page/Function). The growth estimate $|p(z)| \to \infty$ as $|z| \to \infty$ (dominated by the leading term) shows $1/p$ is bounded. [Liouville's theorem](/theorems/346) then forces $1/p$ to be constant, contradicting $\deg p \geq 1$. [/proofplan] [step:Assume $p$ has no root and form the entire function $g = 1/p$] Let $p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_0$ with $a_n \neq 0$ and $n \geq 1$. Suppose for contradiction that $p(z) \neq 0$ for all $z \in \mathbb{C}$. Then \begin{align*} g: \mathbb{C} &\to \mathbb{C} \\ z &\mapsto \frac{1}{p(z)} \end{align*} is well-defined and entire (a quotient of entire functions with non-vanishing denominator). [/step] [step:Show $|p(z)| \to \infty$ as $|z| \to \infty$ to bound $g$ outside a large disc] For $z \neq 0$, factor out the leading term: \begin{align*} p(z) = a_n z^n \left(1 + \frac{a_{n-1}}{a_n z} + \cdots + \frac{a_0}{a_n z^n}\right). \end{align*} The parenthetical expression converges to $1$ as $|z| \to \infty$. Precisely, there exists $R > 0$ such that for $|z| \geq R$: \begin{align*} \left|\frac{a_{n-1}}{a_n z} + \cdots + \frac{a_0}{a_n z^n}\right| \leq \frac{1}{2}, \end{align*} so $|p(z)| \geq |a_n| |z|^n / 2 \geq |a_n| R^n / 2 \geq 1$ (after enlarging $R$ if necessary). Therefore $|g(z)| \leq 1$ for all $|z| \geq R$. [/step] [step:Bound $g$ on the compact disc and apply Liouville's theorem] On the compact set $\overline{B(0, R)}$, the function $g$ is continuous (since $p$ has no zeros), hence bounded: $|g(z)| \leq K$ for some $K > 0$ and all $|z| \leq R$. Combining: $|g(z)| \leq \max(1, K)$ for all $z \in \mathbb{C}$, so $g$ is a bounded entire function. By [Liouville's theorem](/theorems/346), $g$ is constant, hence $p = 1/g$ is constant. This contradicts $n \geq 1$. [/step]