Finite field, every element is a square implies char equal 2

If $F$ is a finite field such that every element is a square, why must $char(F)=2$?

If every element is a square, then the map $x\mapsto x^2$ is a surjection from $F$ to itself. Since $F$ is a finite field, the map is a bijection, which means that since $-1$ and $1$ have the same image, then $-1=1$, hence the characteristic is $2$.

In fact, the converse holds as well. Suppose $F$ is a finite field of characteristic $2$: then it has $2^n$ elements, for some $n$, hence the multiplicative group of nonzero elements is $F$ of order $2^n-1$, which is odd. In particular, there is no element of order $2$ by Lagrange's theorem. The map $f\colon F-\{0\}\to F-\{0\}$ given by $f(x)=x^2$ is a group homomorphism whose kernel consists of all elements of exponent $2$; by the comment just made, the kernel is trivial, hence $f$ is one-to-one and hence onto, so every nonzero element of $F$ is a square. And of course, $0$ is a square.

Another way: all elts of a finite field of size $2n\!+\!1$ are squares $\Rightarrow \color{#0a0}{x^{n}\! - 1} $ has $2n$ roots (all elts $\ne\! 0$), contra a polynomial over a field has no more roots than its degree, by $\,a = b^2\Rightarrow \color{#0a0}{a^n} = b^{\color{#c00}{2n}} \color{#0a0}{= 1},\,$ by Lagrange, since its multiplicative group has size $\,\color{#c00}{2n}$ (i.e. Euler's Criterion when size is prime).