About zeros of a polynomial I wonder why the polynomial $x^p-x$ has $p$ distinct zeros in $\mathbb Z_p$ for any prime $p$, i.e. $x^p-x=x(x-1)\cdots(x-p+1)$.
Do I need to expand the polynomial in order to get the conclusion?
 A: This will help: See Fermat's Little Theorem.
Added: ...as I see @anon suggested!
Point being: You don't need to expand the polynomial!
A: Use little Fermat; or Freshman's Dream (a.k.a. Frobenius map) $\rm\color{#C00}{\,applied\,}$ below
$$\rm mod\ p\!:\,\ f(x)=x^p\!-x\:\Rightarrow\:f(x\!+\!1) = \color{#C00}{(x\!+\!1)^p}\!-x\!-\!1 \,\equiv\, \color{#C00}{x^p\!+1^p}\!-x\!-\!1 \,\equiv\, x^p\! - x = f(x)$$
Hence $\rm\ f(x\!+\!1)\equiv f(x)\ $ so $\rm\ f(p\!-\!1)\equiv f(p\!-\!2)\equiv \cdots \equiv f(1)\equiv f(0)\equiv 0.$
A: Since it is tagged ring theory, I assume that OP already knows some group theory.
As $p$ is a prime, every number in {$1,\cdot\cdot\cdot, p-1$} is prime to $p$, hence, by Bézout identity, the set $\mathbb {Z/pZ}=${$1,\cdot\cdot\cdot,p-1$} forms a group under multiplication.
Now this group has order $=p-1$, so every number $x\in \mathbb {Z/pZ}$ satisfies $x^{p-1}\equiv 1\pmod p^{Notice}$, by Lagrange. And this is equivalent with your result.
Notice: One could also see this directly by using Petit Euler, if there is such a name.
