= Solution
Let $P(x)$ denote the polynomial under consideration. The change $x\mapsto x+1$ is an automorphism of $\mathbb Q[x]$ and preserves degrees, so it preserves reducibility. Its transformed polynomial is
$$
P(x+1)=\frac{(x+1)^p-1}{x}
=\sum_{j=1}^p\binom pjx^{j-1}.
$$
It is monic; every other coefficient is divisible by $p$, because $p$ is prime and $1\leq j<p$; and its constant coefficient is $p$, not divisible by $p^2$. The <Eisenstein criterion> proves that $P(x+1)$ is irreducible, and therefore \b[$P$ is irreducible]. The argument also includes $p=2$, when the polynomial is linear. This is the prime-index case of an <irreducible polynomial> arising as a <cyclotomic polynomial>.
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