proof that $\frac{x^p - 1}{x-1} = 1 + x + \dots + x^{p-1}$ is irreducible I am reading the group theory text of Eugene Dickson.  Theorem 33 shows this polynomial is irreducible
$$ \frac{x^p - 1}{x-1} = 1 + x + \dots + x^{p-1} \in \mathbb{Z}[x]$$
He shows this polynomial is irreducible in $\mathbb{F}_q[x]$ whenever $p$ is a primitive root mod $q$.
By Dirichlet's theorem there are infinitely many primes $q = a + ke$, so this polynomial is "algebraiclly irreducible", I guess in $\mathbb{Q}[x]$.

Do you really need a strong result such as the infinitude of primes in arithmetic sequences in order to prove this result?  Alternative ways of demonstrating this is irreducible for $p$ prime?
COMMENTS Dirichlet's theorem comes straight out of Dickson's book.  I am trying to understand why he did it.  Perhaps he did not know Eisenstein's criterion.  It's always good to have a few proofs on hand.
Another thing is that Eisenstein's criterion is no free lunch since it relays on Gauss lemma and ultimately on extending unique factorization from $\mathbb{Z}$ to $\mathbb{Z}[x]$.

To clarify an issue that came up in the comments, by Gauss's lemma a monic polynomial is irreducible over $\mathbb{Q}[x]$ iff it's irreducible over $\mathbb{Z}[x]$, and by reduction $\bmod p$, if a polynomial is irreducible $\bmod p$ for any particular prime $p$, then it's irreducible over $\mathbb{Z}[x]$. So once you know that the polynomial is irreducible $\bmod q$ for any prime $q$ which is a primitive root $\bmod p$, it suffices to exhibit one such prime. 
Unfortunately, as far as I know this is no easier than Dirichlet's theorem. Among other things, it's a straightforward exercise to show that Dirichlet's theorem is equivalent to the assertion that for any $a, n, n \ge 2$ such that $\gcd(a, n) = 1$ there is at least one prime (rather than infinitely many primes) congruent to $a \bmod n$. 
Regarding your last comment, Eisenstein's criterion is still much, much, much easier to prove than Dirichlet's theorem, even with all of the details filled in. 
