I want to prove that for finite $\mathbb{F}$, there exists $n\in\mathbb{Z}^+$ such that $$ \underbrace{1+1+...+1}_{\text{n times}} = 0 $$
Proof: by the field axioms, there exists $1\in\mathbb{F}$. Also, by closure, $1+1\in\mathbb{F}$. We can continue this and construct a set $X = \{1,1+1,1+1+1,...\}$. Since $\mathbb{F}$ is finite, let $|\mathbb{F}|=k$. We construct $X$ so that it has $k+1$ elements. Then, by pigeonhole principle, there must exist $l,m\in\mathbb{F}$ such that $$ l = \underbrace{1+1+...+1}_{\text{a times}}=\underbrace{1+1+...+1}_{\text{b times}} = m $$ WLOG $a>b$. Then $$ \underbrace{1+1+...+1}_{\text{a-b times}}=0 $$ which is the result of the theorem.
I'm sure this is valid but I'd like to check. Any suggestions/errors?