TITLE: Given a prime $p\in\mathbb{N}$, is $A=\frac{x^{p^{2}}-1}{x^{p}-1}$ irreducible in $\mathbb{Q}[x]$?
QUESTION [6 upvotes]: If $p \in \mathbb{N}$ is a prime, is $\displaystyle A=\frac{x^{p^{2}}-1}{x^{p}-1}$ irreducible in $\mathbb{Q}[x]$?   
I don't think it is. If somebody sees a contradiction, I would be glad to see it. 
The link of Qiaochu links to the Eisenstein criterion. 
There it is written that: 
cyclotomic polynomials can be obtained by dividing the polynomial $x^{p}-1$ (in this case $x^{p^{2}}-1$) by $x-1$ (in this case $x^{p}-1$, which is it's obvious root. 
Then the article makes a substitution so the criterion can be applied: by substituing x+1 for x this gives : $((x+1)^{p}-1)/x = x^{p-1} + (p nCR p-1)x^{p-2} + .... + (p nCR 1) $. The coefficients are divisible by p  because of the properties of binomial coefficients, and therefore not divisible by $p^{2}$. 
How to find the substitution so Eisenstein criterion can be applied? 
I am very thankful for any insights.

REPLY [3 votes]: Here is a proof based on the Eisenstein criterion, without using the general result about the irreducibility of cyclotomic polynomials.
For the $p^2$-th cyclotomic polynomial $A = \frac{x^{p^2} - 1}{x^p - 1}$, we are going to show that $A(x+1)$ is an Eisenstein polynomial for the prime $p$.
Substituting $x+1$ into the equation $(x^p - 1)A(x) = x^{p^2} - 1$ yields (after the application of the binomial theorem and division by $x$)
$$\left(\sum_{i=0}^{p-1}\binom{p}{i+1}x^i\right)\cdot A(x+1) = \sum_{i=0}^{p^2 - 1}\binom{p^2}{i+1}x^i.$$
Comparison of the constant coefficients shows that $pa = p^2$, where $a$ denotes the constant coefficient of $A(x+1)$. So $a = p$.
Furthermore, reading the above equation modulo $p$ we get
$$x^{p-1} \cdot A(x+1) \equiv x^{p^2 - 1}\mod p$$
and therefore
$$A(x+1) \equiv x^{p^2 - p}\mod p.$$
So $A(x+1)$ is indeed an Eisenstein polynomial.