Question on proof about finite fields mentioned on wikipedia In the first paragraph of wikipedia:Finite fields they write

The identity
  $$  (x + y)^p = x^p + y^p $$
  is true (for every $x$ and $y$) in a field of characteristic $p$.
For every element $x$ in the prime field $GF(p)$, one has $x^p = x$ (This is an immediate consequence of Fermat's little theorem, and this may be easily proved as follows: the equality is trivially true for $x = 0$ and $x = 1$; one obtains the result for the other elements of $GF(p)$ by applying the above identity to $x$ and $1$, where $x$ successively takes the values $1,2,\ldots, p-1$ modulo $p$. This implies the equality
  $$
 X^p - X = \prod_{a\in GF(p)} (X - a)
$$
  for polynomials over $GF(p)$.

I do not understand what they are doing in the last cited paragraph, they want to apply $(x + 1)^p = x^p + 1$ for $x = 1,2,\ldots, p-1$ modulo $p$, but how does this gives the result $x^p = x$ or that $X^p - X = \prod_{a\in GF(p)} (X - a)$; I do not see what is happening here?
 A: For the equality $x^p = x$, they assume that the statement is proven for some $a\in GL(p)$, and then prove it for $a+1$ by writing $(a+1)^p = a^p + 1^p = a+1$. So it is a form of induction. 
For the second part, let's write for the moment $f(X) = X^p -X$ and $g(X) = \prod_{a\in GL(p)} (X-a)$. We want to prove that $f=g$. At least we know that $f$ and $g$ have the same degree and the same leading coefficient. From the definition of $g$ we know that every element of $GL(p)$ is a root of $g$, with multiplicity one, and these are all the roots of $g$. On the other hand, because of the first paragraph, every element of $GL(p)$ is also a root of $f$, and since $f$ has degree $p$, all these roots must have multiplicity 1 as well. Therefore, $f$ and $g$ have the same roots (including multiplicity) and the same leading coefficient, so $f=g$. 
A: For the first part, we apply induction on $x$ to get $(x+1)^p = x^p + 1^p = x+1$ as desired. The second equation is an equivalence of two polynomials over $\mathbb{Z}/p\mathbb{Z}$. Since $\mathbb{Z}/p\mathbb{Z}$ is a domain, a polynomial is zero at $a$ iff it's divisible by $(x-a)$. The polynomial $x^p-x$ is zero at each $x \in \mathbb{Z}/p\mathbb{Z}$ (since $x^p = x$) hence it is divisible by $\Pi (x-a_i)$. Considering degrees, it must equal $\Pi (x-a_i)$.
