Eisenstein Criterion with a twist As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a prime, $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$?
 A: Here is the standard proof, mentioned by Arturo and Gerry
$$P(x)=\frac{x^p-1}{x-1} \,.$$
Then 
$$P(x+1) = \frac{(x+1)^p-1}{x} =\frac{ \sum_{k=0}^p \binom{p}{k}x^p -1}{x}=x^{p-1} \sum_{k=1}^{p-1} \binom{p}{k}x^{k-1} \,.$$
It is trivial to show that $p| \binom{p}{k}$ for all $1 \leq k \leq p$ and that $p^2 \nmid  \binom{p}{1}$, thus Eisenstein Criteria can be applied.
Since $P(x+1)$ is irreducible, we have $P(x)$ is also irreducible...
A: Writing down Arturo's answer: if $\,f(x):=x^{p-1}+x^{p-2}+...+x+1\,$ , then
$$f(x+1)=(x+1)^{p-1}+(x+1)^{p-2}+...+(x+1)+1=$$
$$=\sum_{k=0}^{p-1}\binom {p-1}{k}x^k+\sum_{k=0}^{p-2}\binom{p-2}{k}x^k+\ldots+\sum_{k=0}^1\binom{1}{k}x^k+\sum_{k=0}^0\binom{0}{0}x^k$$
so the coefficient of $\,x^j\,\,,\,0\leq j\leq p-1\,$ , in the above is:
$$\sum_{m=1}^{p-j}\binom{p-m}{j}=\sum_{m=0}^{p-1-j}\binom{j+m}{j}\,\text{, and, of course, }\,\,\binom{t}{r}=0\,\,\text{whenever}\,\,r>t $$
We shall prove by induction on $\,j\,$ that the above coefficient is divisible by $\,p\,$:
$$ ==\,\,\text{For}\,\,j=0:\,\,\sum_{m=0}^{p-1}\binom{m}{0}=\sum_{m=0}^{p-1}\,1=p$$
Assume now for $\,l\leq j-1\,$ and we shall prove for $\,l=j\,$ . We use the basic identity
$$\binom{a+1}{b+1}=\binom{a}{b+1}+\binom{a}{b}\,:$$
$$\sum_{m=0}^{p-1-k}\binom{m+j}{j}=\sum_{m=0}^{p-1-j}\binom{m+j-1}{j}+\sum_{m=0}^{p-1-j}\binom{m+j-1}{j-1}=$$
$$\sum_{m=0}^{p-j}\binom{m+j-1}{j}-\binom{p-1}{j}+\sum_{m=0}^{p-j}\binom{m+j-1}{j-1}-\binom{p-1}{j-1}\;\;\;(**)$$
The third sum above is divisible by $\,p\,$ by the inductive hypthesis, also:
$$\binom{p-1}{j}+\binom{p-1}{j-1}=\binom{p}{j}=p\cdot\frac{(p-j+1)(p-j+2)\cdot\ldots\cdot(p-1)}{j!}$$
is divisible by $\,p\,$ , and making the substitution $\,j\to t-1\,$ in the first sum in (**) we get
$$\sum_{m=0}^{p-j}\binom{m+j-1}{j}\to\sum_{m=0}^{p-t+1}\binom{m+t-2}{t-1}=\sum_{m=0}^{p-t}\binom{m+t-1}{t-1}$$
Which again is the same sum as in $\,(**)\,$ and the inductive hypothesis again gives divisibility by $\,p\,$, getting thus that the whole expression (**) is divisible by $\,p\,$
