If a finite field has characteristic 2 why is every element a square? 
If a finite field has characteristic 2 why is every element a square?

I've attempted this problem by calculating the square root as follows: $\mathbb{F}$ has $q = 2^m$ elements, so if $a \in \mathbb{F}$ then $a = a^q = (a^{2^{(n-1)}})^2$. Since, $a \in \mathbb{F}$ is a square elements if there exist $b \in \mathbb{F}$ s.t $b^2 = a$, I believe the next step would be to show that the square root $b = a^{2^{(n-1)}} \in \mathbb{F}$. Is this reasonable or should I take a different approach to the problem (and if so what approach?)?
 A: Consider the function $f : \Bbb F \to \Bbb F$ given by $f(x) = x^2$.
Then, $$f(x) = f(y) \iff x^2 = y^2 \iff x^2 - y^2 = 0 \color{red}{\iff} (x - y)^2 = 0 \iff x - y = 0.$$
The $\color{red}{\iff}$ is true since the field has characteristic $2$. Thus, $f$ is actually injective. An injective map from a finite set to itself is always surjective and thus, every element of $\Bbb F$ is a square.

In general, the above argument shows that if $\Bbb F$ is a finite field with characteristic $p$, then $x \mapsto x^p$ is a bijection. In fact, this is actually an isomorphism since $(x + y)^p = x^p + y^p$.

Note that finiteness is crucial. For example, consider the field $\Bbb F = \Bbb F_{2}(T)$ of rational functions over $\Bbb F_{2}$ in the indeterminate $T$. In this case, the map above is an injection but the image is strictly smaller than $\Bbb F$. For example, $T$ is not in the image.
A: You can do it like this.
If $a^2=b^2$ then $a=\pm b$, but in characteristic $2$ we have $b=-b$ so that if $a^2=b^2$ then $a=b$.
Now square all the $n$ distinct elements of your finite field. You get $n$ different answers. And since there are only $n$ elements in the field each of them must appear in the list of squares.

You can also do it by noting that the multiplicative group of non-zero elements has odd order, and every element of a group of odd order is a square. Working with $q$ in your case is a simple case of this underlying fact.
