Consider the polynomial:
$$(X+1)^{p(p-1)} + (X+1)^{p(p-2)} + \cdots + (X+1)^p + 1$$
In Serre's local fields, he says that this polynomial is clearly Eisenstein, by noting that its constant term is $p$, and mod $p$ it is equal to $X^{p(p-1)}$.
It's clear that it's constant term is $p$, but it is far less clear to me that mod $p$ it is just $X^{p(p-1)}$. I've reduced it to showing that for any $0 < i < p$, the sum $\binom{p-1}{i} + \binom{p-2}{i-1} + \cdots + \binom{p-1-i}{0}$ is divisible by $p$, but this is hardly obvious to me.
Is there a better way to see that all the non-leading terms vanish mod $p$?
Use $$u^p- 1 = (u - 1 )(u^{p-1} + \cdots +1),$$ and $$ (x+1)^p \equiv x^p + 1 \pmod p,$$along with the fact that ${\mathbb F}_p[x]$ is a UFD.
Edit:
As Darij points out below, one does not need to use the UFDhoodshipness of ${\mathbb F}_p[x]$. For any (commutative) ring $R$, multiplication by $x$ is (visibly!) injective on the polynomial ring $R[x]$:
Set $$ q(x) = (x+1)^{p(p-1)}+ \cdots + 1. $$ The argument sketched above (the 2 equalities) gives the equality in $ {\mathbb F}_p[x]$ $$ x^{p^2} = x^p q(x).$$ Therefore, by the injectivity by multiplication by $x$ in the polynomial ring, $$q(x) = x^{p^2-p}.$$