If $p$ is a prime prove that $x^{p-1} - x^{p-2} + x^{p-3}- \cdots -x+1$ is irreducible over $Q$ If $p$ is a prime prove that $x^{p-1} - x^{p-2} + x^{p-3}- \cdots -x+1$ is irreducible over $Q$.
$1$st Attempt: $x^{p-1} - x^{p-2} + x^{p-3}- \cdots -x+1 $
$= x^{p-2}(x-1)+x^{p-4}(x-1)+ \cdots ....x(x-1)+1$
$=(x-1)(x^{p-2}+x^{p-4}+ \cdots+x)+1$
I don't have much idea on how to move forward.
$2$nd Atempt: Using Fermat's theorem : if $a^{p-1} \equiv 1 \mod p~;~(a \ne p)$
Let $x=m/n$ such that $m,n \in Z ~;~\gcd(m,n)=1$. Then if 
$f(x) = x^{p-1} - x^{p-2} + x^{p-3}- \cdots -x+1 $
$n^{p-1}f(x) = g(x)=m^{p-1} -n.m^{p-2}+n^2.m^{p-3}-\cdots-n^{p-2}m+n^{p-1}~~\in~~Z[x]$
I am not able to see how fermat's theorem can be used cleverly enough to make a difference here.
$3$rd Attempt: $\mod p$ irreducibility test and Eisenstein's Criterion, which can be brought into the required format only perhaps by substituting $g(x+a)$, again tedious.
Thank you for your help.

Hint $ $ If suffices to prove irreducible $\rm\,f(x) = g(-x) = (x^p-1)/(x-1).\,$ Eisenstein's Criterion applies to polynomials that have the form $\rm\, f \,\equiv\, x^n\pmod p.\,$ $\rm\,f\,$ is not of that form, but it's close
$$\rm f\, = \dfrac{x^p-1}{x-1} \equiv \dfrac{(x-1)^p}{x-1}\equiv (x-1)^{p-1}\pmod p$$
Eisenstein's Criterion may apply if you can find a map $\ \sigma\ $ that sends $\rm\ (x-1)^{p-1}\ $ to a power of $\rm\ x\ $ and, further, $\ \sigma\ $ preserves factorizations $\rm\ \sigma(gh)\ =\ \sigma g\cdot \sigma h\ $ (so we can pullback the irreducibility of $\rm\ \sigma\:f\ $ to $\rm\:f\:\!).\:$ For this, it suffices to find a  find an automorphism $\:\sigma\:$ on $\rm\, \Bbb Z[x]\, $ that shifts $\rm\:x\!-\!1\to x.\, $ Any ideas?
