This is part of 'Abstract Algebra, Paul Garrett, 243-244p'
Recall that $Φ_{p^e}(x) = Φ_p(x^{p^{e-1}}) = \frac{x^{p^{e}}-1}{x^{p^{e-1}}-1} $
First, we check that $p$ divides all but the highest-degree coefficient of $f(x)$. To do so, map everything to $\mathbb{F}_p[x]$, by reducing coefficients modulo p. For $e ≥ 1$ $(x + 1)^{p^{e-1}} = x^{p^{e-1}} + 1$ ($mod$ $p)$
Therefore, in $\mathbb{F}_p[x]$ $f(x) = Φ_p((x+1)^{p^{e-1}}) = \frac{(x+1)^{p^{e}}-1}{(x+1)^{p^{e-1}}-1}$
$ = ((x+1)^{p^{e-1}})^{p-1}+((x+1)^{p^{e-1}})^{p-2}+...+((x+1)^{p^{e-1}})+1$
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$ = (x^{p^{e-1}}+1)^{p-1}+(x^{p^{e-1}}+1)^{p-2}+...(x^{p^{e-1}}+1)+1$
$ = \frac{(x^{p^{e-1}}+1)^p-1}{(x^{p^{e-1}}+1)-1}$
$ = \frac{x^{p^{e-1}}+1-1}{x^{p^{e-1}}}$
$ = \frac{x^{p^e}}{x^{p^{e-1}}}$
$ = x^{p^{e-1}(p-1)}$
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in $\mathbb{F}_p[x]$ Thus, all the lower coefficients are divisible by $p$. To determine the constant coefficient of $f(x)$, again use
$Φ_{p^e}(x) = Φ_p(x^{p^{e-1}})$
to compute
constant coefficient of $f = f(0) = Φ_{p^e} (1) = Φ_p(1^{p^{e-1}}) = Φ_p(1) = p$
as in the prime-order case. Thus, $p^2$ does not divide the constant coefficient of f. Then apply Eisenstein’s criterion and Gauss’ lemma to obtain the irreducibility
I don't understand ........... part
Based on the assumption of $p = 5, e = 2 $ from Wolfram Alpha, $\frac{(x + 1)^{25} - 1}{(x + 1)^5 - 1} = ((x+1)^5)^4+((x+1)^5)^3+((x+1)^5)^2+(x+1)^5+1$
$≠ (x^5+1)^4+(x^5+1)^3+(x^5+1)^2+(x^5+1)+1$