Let $x=y+1$. Then we get \[\frac{(y+1)^{p}-1}{y}= y^{p-1}+\binom{p}{1}y^{p-2}+.....+\binom{p}{p-1}\] All the binomial coefficients are divisible by $p$, $\binom{p}{p-1}=p$ and therefore cannot be divisible by $p^{2}$, and the leading coefficient is 1, so it follows from Eisenstein's Criterion (see Q5).

Proof that cyclotomic polynomials are irreducible: Lemma: If $\Phi_n(x) = f(x)g(x)$ is a factorisation of the $n$-th cyclotomic polynomial (over $\mathbb{Q}$) and $p$ a prime that doesn't divide $n$, then $f(\zeta)=0$ implies $f( \zeta ^p)=0$. Proof: Assume that $\zeta ^p$ is not a root of $f$, so it's a root of $g$. So consider the polynomials modulo $p$ (this is possible since all coefficients are integers) to get (by writing out the polynomial in term's of binomial coefficients using the binomial theorem) that $f(\zeta ^p) \equiv f(\zeta)^p \equiv 0 \mod p$ (think of $\zeta \mod p$ as some number 'added' to the 'normal' residue classes $\mod p$). But since also $g(\zeta^p)\equiv 0 \mod p$, there is some root occuring twice $\mod p$. But this is impossible, since the derivative of $x^n-1 \mod p$ is $nx^{n-1}$ and so different from zero for $z\not\equiv 0 \mod p$ (it's the same like in the real numbers: you only can have a double root if the polynomial and it's derivative are both zero, the proof for that isn't that hard). This contradiction implies that $f(\zeta ^p)=0$. To prove the theorem now, simple take some representing integer $k$ of a prime residue class $\mod n$ and factor it into primes (not dividing $n$). Now use the lemma inductively on the factors of $k$ to prove that with $\zeta$ being a root of $f$ also $\zeta^k$ is necessary a root and by this all primitive $n$-th roots of unity are roots of $f$. But this shows that $f=\Phi_n$, so the cyclotomic polynomial is irreducible.

Ilthigore wrote: Let $x=y+1$. Then we get \[\frac{(y+1)^{p}-1}{y}= y^{p-1}+\binom{p}{1}y^{p-2}+.....+\binom{p}{p-1}\] All the binomial coefficients are divisible by $p$, $\binom{p}{p-1}=p$ and therefore cannot be divisible by $p^{2}$, and the leading coefficient is 1, so it follows from Eisenstein's Criterion (see Q5). Do you have a paper where Eisenstein's Criterion is well explained?

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=489&t=150950