Consider a finite vector space V over a field F where |V| > 1. Is F always finite?
My proof: Suppose for a contradiction that F was infinite. Since V is closed under scalar multiplication any element from our field, say c, multiplied by 1 will give us back c, which is in V. Since F has infinitely many elements this implies V will also have infinitely many elements. This is a contradiction since V is finite. Hence F must also be finite.
Suppose that $F$ is not finite. Let $v\neq 0, v\in V$, consider the map $f:F\rightarrow V$ defined by $f(x)=xv$, there exists $x,y$ not zero, $x\neq y$ such that $f(x)=f(y)$ this implies that $(x-y)v=0$, by multiplying by the inverse of $(x-y)$ we deduce that $v=0$ contradiction.