Let $m(X) := \frac{X^p-1}{X-1}$ and $n(X) := m(X^p)$. I have shown that $m$ is irr. over $\mathbb Q$. Now I want to show that this is true for $n(X)$, too. I know that $$ n(X+1)((X+1)^p-1)= (X+1)^{p^2}-1 $$ Working modulo $p$ I get that $$ n(X+1)X^p \equiv X^{p^2} \mod p $$ which means that $p$ divives each coefficient of $n(X+1)$ and not the leading coefficient. But I do not know how to show that $p^2 \nmid a_0$ where $a_0$ is the constant term of $n(X+1)$.
Pleas help :)
Turning my comment into an answer: write $m(X) = \sum_{i=0}^{p-1}X^i = 1+X+X^2+\ldots+X^{p-1}$. Then $n(X+1) = m\left((X+1)^p\right) = 1+(X+1)^p+(X+1)^{2p}+\ldots+(X+1)^{p(p-1)}$. Now just use the binomial theorem to find the constant term of each of the terms of this sum.