$(u+1)(u+2)\cdots(u+p)=u^p-u$ Let $E$ be a field with characteristic $p>0$. How should I prove that
$$(u+1)(u+2)\cdots(u+p)=u^p-u$$
I can verify this equality for small $p=2,3,5$. Is there a way to prove this result in general? This question actually arises from the following exercise:

Let $k$ be a field of characteristics $p>0$, and let $E/k$ be a
    Galois extension having a cyclic Galois group
    $G=\langle\sigma\rangle$ of order $p$. Prove that there is an element $u\in E$ with $\sigma(u)-u=1$. Prove that $E=k(u)$ and that there is $c\in k$ with $irr(u,k)=x^p-x-c$.

I need to prove that $irr(u,k)=x^p-x-c$ for some $c\in k$. Since all roots of $irr(u,k)$ must be $u+i$, $0\leq i<p$, $c=(u+1)(u+2)\cdots (u+p)=N(u)$, the norm of $u$ that lies in $k$. Since $x^p-x-c$ is irreducible in $k[x]$, all I need to do is to prove that $u$ is a root of $x^p-x-c$; that is, $c=u^p-u$. That is why I ask the question.
Any help?

Hint: The roots of the polynomials $X^p-X$ and $(X+1)(X+2)\dots(X+p)$ are…
