Existence of elements of even order in a field with characteristic 2

Question

I've read this statement in a presentation slide, but it isn't obvious to me on why this is true:

Forgetting about BCH codes, the question is: if an alement $\beta$ has even order ($2k$ is always even for $k \in \mathbb{N}$), why can't such an element exist in a field with characteristic 2 (e.g. in $\mathbb{F}_2$)? I know that the characteristic of a field is defined as

$$char(F) = k \Leftrightarrow \underbrace{1 + 1 + ... + 1}_{k\text{ times}} = 0 $$ and the $ord(\beta)$ is defined as the smallest $a$ for which the statement $\beta^a = 1$ holds true, but I can't seem to wrap my head around why the upper statement must be true. An explanation of this is highly appreciated.

Answer 1

In a finite group, the order of an element divides the order of the group. A finite field of characteristic $2$ has $2^n$ elements for some positive integer $n$, so its multiplicative group has odd order.

Answer 2

To see this directly: If there is an element of order $2k$, then there is an element of order $2$.

But $x^2-1 = (x-1)^2$ in a field of characteristic $2$, so there are no nontrivial square roots of $1$.