TITLE: $p^{th}$ roots of a field with characteristic $p$
QUESTION [14 upvotes]: This is problem 10.9 from the book "Error-Correcting Codes and Finite Fields by Oliver Pretzel".
The Question:

Show that in a field of characteristic $p$, any element $\alpha$ has at most one $p$-th root $\beta$ (i.e., an element $\beta\in F$ with $\beta^p = \alpha$). Show further that if $F$ is finite, then every element has exactly one $p$-th root

This my attempt at the second part of the question.
From Fermat's little theorem $\beta^{{p^n}-1} = 1$, where $p^n$ is the size of the field. Now multiplying both sides by $\beta$ we get $\beta^{p^n} = \beta$.
If there is $p$ elements then $n=1$ and we can see this is true for any non-zero element. 
For the general case, take the $p$-th root of both sides $\beta^{p^{n-1}} = \beta^{1/p}$ and we know from the multiplicative properties of a field if $\beta$ is a non zero element of the field then any multiple will be.
The first part of the question I'm not sure where to begin.
Any help would be appreciated.

REPLY [10 votes]: If $a^p = b^p$ then $a^p-b^p = (a-b)^p=0$, and since you are in a field this implies $a=b$.  This shows that for a field of characteristic $p$ the map $a \to a^p$ is always injective, and an injective map from a finite set to itself is automatically bijective.