 Solution

The Eisenstein criterion says that a primitive integer polynomial $P(x)=a_m{x}^m+\cdots +a_0$ is irreducible over $\mathbb{Q}$ if some prime number $p$ divides every $a_j$ for $j<m$, does not divide $a_m$, and $p^2$ does not divide $a_0$. By Gauss lemma for polynomials, this gives irreducibility in $\mathbb{Z}[x]$ as well.

For prime $n=p$, translate the variable in $P_p(x)=(x^p-1)/(x-1)$:Its leading coefficient is one, all other coefficients are divisible by $p$, and its constant coefficient is exactly $p$. The Eisenstein criterion proves this translated polynomial irreducible. Since $x\mapsto x+1$ is an invertible substitution in $\mathbb{Z}[x]$, $P_p$ is irreducible too.

If $n=ab$ with $a,b>1$, thenBoth factors have positive degree and integer coefficients, proving reducibility. Consequently $\boxed{1+x+\cdots +x^{n-1}\text{ is irreducible exactly when }n\text{ is prime}.}$