Multiplicative group of an infinite field is not cyclic

Ok here is the characteristic 2 case:

Assume $k$ is an infinite field of characteristic $2$ with a cyclic multiplicative group. Note that any element of an algebraic extension of $\mathbb{F}_2$ has finite multiplicative order, so this implies that every element of $k-\{0,1\}$ must be transcendental.

Next let $x$ be a generator for the multiplicative group, which exists as we are assuming it is cyclic. Consider the element $1+x$ of our field. It is nonzero and therefore equal to some power of $x$ since $x$ is a generator. But then $1+x=x^n$ for some $n$, so $x$ is algebraic over $\mathbb{F}_2$, contradicting the above claim.

I was studying this for a Galois Theory course and also bumped into this question, and actually for fields of characteristic different from 2 there is a really surprisingly simple proof (spoiler, this will happen because in characteristic $2$ we get $-1=1$).
So suppose $k$ is an infinite field such that char $k \neq 2$ and there is $x$ with $\langle x \rangle = k^*$. This implies that there is an $n \in \mathbb{N}$ (different than $0$ because $-1 \neq 1$) such that $x^n = -1$, which implies that $x^{2n} = 1$ and so $k^*$ is finite, contradiction.