= Solution
The <Eisenstein criterion> says that a primitive integer <polynomial> $P(x)=a_mx^m+\cdots+a_0$ 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 $P_p(x)=(x^p-1)/(x-1)$:
$$
P_p(x+1)=\frac{(x+1)^p-1}{x}=\sum_{j=1}^p\binom pj x^{j-1}.
$$
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 $a,b>1$, then
$$
P_n(x)=(1+x+\cdots+x^{a-1})(1+x^a+\cdots+x^{(b-1)a}).
$$
Both factors have positive degree and integer coefficients, proving reducibility. Consequently $\boxed{1+x+\cdots+x^{n-1}\text{ is irreducible exactly when }n\text{ is prime}.}$
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