Multiplicative group of an infinite field is not cyclic Question.

Prove  that  the  multiplicative  group  of any  infinite  field  can  never  be  cyclic .

$\mathbb R$, $\mathbb Q$, $\mathbb C$ are some infinite fields whose multiplicative groups are not cyclic, I know.
I  need  some  lead  as  to  how  to  begin  the  proof.
Sorry  for  the  lack  of   work  on  my  part (I'm  clueless)  and  any  help  is  appreciated.
 A: 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.
A: 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.
