Let $p$ be prime and consider the polynomial $$f(x)=x^{p-1}+x^{p-2}+\dots+x^2+x+1 $$ Prove that $f(x)$ is irreducible
Hint) May use without proof that $p|\binom {p} {a}$ with $a: 1\leq a \leq p-1 $
Road plan is to use Eisenstein criterion for irreducibly in $\mathbb{Q}[x]$. We need to shift first of all, $$f(x+1)=(x+1)^{p-1}+(x+1)^{p-2}+\dots+(x+1)^2+(x+1)+1 $$ Using Binomial formula to expand $$\begin{aligned} &(x+1)^{p-1}=\sum^{p-1}_{k=0}\binom{p-1}{k}x^{k} &=\binom{p-1}{0}x^{0}+\binom{p-1}{1}x^{1} +\dots+\binom{p-1}{p-2}x^{p-2}+\binom{p-1}{p-1}x^{p-1}\\ &(x+1)^{p-2}=\sum^{p-2}_{k=0}\binom{p-2}{k}x^{k} &=\binom{p-2}{0}x^{0}+\binom{p-2}{1}x^{1} +\dots+\binom{p-2}{p-3}x^{p-3}+\binom{p-2}{p-2}x^{p-2}\\ \end{aligned} $$ $$\vdots$$ $$(x+1)^1=x+1\\ (x+1)^0=1$$ Missing important steps $$f(x+1)=x^{p-1}+\binom{p}{p-1}x^{p-2}+\dots+\binom{p}{2}x+\binom{p}{1}$$ we know that $p|\binom{p}{p-1},\binom{p}{p-2},\dots,\binom{p}{2},\binom{p}{1}$ and $p\nmid 1$ & $p^2 \nmid (\binom{p}{1}=p)$
$\therefore$ thanks to eisentien's theorem it is irreducible in $\mathbb{Q}[x]$
Could use some hints on the missing steps
Another road is to use this Thm
$p(x)=a_n x^n+\dots+a_1 x+a_0 \in \mathbb{Z}[x]$
p is a prime where $p \nmid a_n$
consider $\phi: \mathbb{Z}[x] \to \mathbb{Z}_p[x]$
if $\phi(f)$ is irreducible $\Rightarrow$ f is irreducible
We know that $p$ divides the binomial coefficients for $1 \leq k \leq p-1$, and that there will be $p-1$ 1's plus one more left over from expanding out everything. So $p$ will divide your $a_0$ and all your coefficients except the first term. You've got it from here.