TITLE: Eisenstein Criterion with a twist
QUESTION [6 upvotes]: As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a prime, $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$?

REPLY [12 votes]: Here is the standard proof, mentioned by Arturo and Gerry
$$P(x)=\frac{x^p-1}{x-1} \,.$$
Then 
$$P(x+1) = \frac{(x+1)^p-1}{x} =\frac{ \sum_{k=0}^p \binom{p}{k}x^p -1}{x}=x^{p-1} \sum_{k=1}^{p-1} \binom{p}{k}x^{k-1} \,.$$
It is trivial to show that $p| \binom{p}{k}$ for all $1 \leq k \leq p$ and that $p^2 \nmid  \binom{p}{1}$, thus Eisenstein Criteria can be applied.
Since $P(x+1)$ is irreducible, we have $P(x)$ is also irreducible...