$f_{p} (x^{p^{e-1}})$ is an irreducible polynomial in $\mathbb{Q}[x]$ for every prime $p$ and every postive integer$e$. For a positive integer $n$ , let 
$$f_{n}(x)= x^{n-1} + x^{n-2} +...+x+1.$$
Then $f_{p} (x^{p^{e-1}})$ is an irreducible polynomial in $\mathbb{Q}[x]$ for every prime $p$ and every postive integer $e$. [True/False].
What I know is that, $f_{p} (x)$ is an irreducible polynomial in $\mathbb{Q}[x]$ for every prime number $p$ .
Any help would be appreciated. Thanks in advance.
 A: The claim is true! You can show that for any prime $p$, and positive integer $e$
$$x^{p^{e+1}} - 1 = (x^{p{^e}} -1)((x^{p{^e}})^{p-1} + ((x^{p{^e}})^{p-2} + \cdots + (x^{p^{e}}) + 1)$$
(to see this more easily, set $u = x^{p^e}$ in into the familiar equation $u^p - 1 = (u-1)(u^{p-1} + \cdots + u + 1)$.
Let $\zeta_{p^{e+1}}$ be a primitive $p^{e+1}$ root of unity. Then, $\zeta_{p^{e+1}}$ is a root of the left hand side of the above equation and so $\zeta_{p^{e+1}}$ is a root of the left hand side. Hence, $\zeta_{p^{e+1}}$ is a root fo $x^{p^e} - 1$ or $(x^{p{^e}})^{p-1} + ((x^{p{^e}})^{p-2} + \cdots + (x^{p^{e}}) + 1$. It is not a root of $x^{p^e} - 1$, it must be a root of $(x^{p{^e}})^{p-1} + ((x^{p{^e}})^{p-2} + \cdots + (x^{p^{e}}) + 1$. The degree of $\zeta_{p^{e+1}}$ over $\mathbb{Q}$ is $\varphi(p^{e+1}) = p^{e+1} - p^e$ (this is the cyclotomic field of $p^{e+1}$ roots of unity). Also, note here that we are using the fact that $p$ is prime to calculate $\varphi(p^{e+1})$.  Thus, the minimal polynomial of $\zeta_{p^{e+1}}$ over $\mathbb{Q}$ is of degree $p^{e+1} - p^e$. 
Note that $(x^{p{^e}})^{p-1} + ((x^{p{^e}})^{p-2} + \cdots + (x^{p^{e}}) + 1$ is a polynomial of degree $p^{e+1} - p^e$ and has $\zeta_{p^{e+1}}$ as a root. Thus, $(x^{p{^e}})^{p-1} + ((x^{p{^e}})^{p-2} + \cdots + (x^{p^{e}}) + 1$ is the minimal polynomial of $\zeta_{p^{e+1}}$ over $\mathbb{Q}$ and consequently irreducible over $\mathbb{Q}$.
