This is a famous result. Let $\alpha$ be any complex roots of $P(x)$. Suppose that $|\alpha| \le 1$, therefore, we have \[ |c_n| = |c_0 \alpha^n + c_1 \alpha^{n - 1} + \dots + c_{n - 1} \alpha | \le |c_0| + |c_1| + \dots + |c_{n - 1} | \]which is a contradiction. Therefore, any complex root of $P(x)$ must satisfies $|\alpha| > 1$. Now, suppose that $P(x)$ is reducible, $P(x) = Q(x) \cdot R(x)$ where $Q, R$ are nonconstant integer polynomials. Therefore \[ c_n = P(0) = Q(0) \cdot R(0) \]Since $|c_n|$ is a prime number, one of $|Q(0)|$ or $|R(0)|$ is equal to $1$. WLOG $|Q(0)| = 1$. Therefore the product of all roots of $Q(x)$ is at most $1$, which is a contradiction.

Sad to see this in an olympiad shortlist as this is extremely well known. In fact, lemmas like Perron's criterion are also common to quite some extent.