= Solution
Put $S=\sum_{j=0}^{n-1}|a_j|$ and choose $R>\max(1,2S/|a_n|)$. For $|z|>R$, the <triangle inequality> gives
$$
\left|\sum_{j=0}^{n-1}a_jz^j\right|\leq S|z|^{n-1}<\frac{|a_n|}{2}|z|^n.
$$
The <reverse triangle inequality> therefore yields
$$
\boxed{|P(z)|>\frac{|a_n|}{2}|z|^n\qquad(|z|>R).}
$$
If $P$ had no <root>, the <function> $h=1/P$ would be an <entire function>. The bound just proved makes $h(z)\to0$ as $|z|\to\infty$. On the remaining closed <disk>, <continuity> and the <extreme value theorem> make $h$ <bounded>. Thus $h$ is a bounded <entire function>, so <Liouville's theorem> makes it constant. Its limit at infinity forces that constant to be zero, impossible for a reciprocal.
Consequently \b[$P$ has a <root> in $\mathbb C$]. This proves the existence part of the <fundamental theorem of algebra>. The invoked <Liouville theorem> follows from the <Cauchy integral formula>: the <Cauchy estimate> $|h'(w)|\leq M/r$ on arbitrarily large <circles> gives $h'=0$ when $h$ is bounded by $M$.
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