Prove that $(R^a)^b = R^{ab}$ for any integers $a,b >= 1$.
A handy fact: The connectivity relation $R^*$ consists of the pairs $(x, y)$ such that there is a path of length at least $1$ from $x$ to $y$ in $R$. Because $R^n$ consists of the pairs $(x, y)$ such that there is a path of length $n$ from $x$ to $y$, it follows that $R^*$ is the union of all the sets $R^n$.
Update: Upon further thought, I think this is a proof to be completed using mathematical induction. The basis step would be $(R^1)^1$ to get $R^{1*1}$ which are both R, hence true. Would the induction step be to prove $(R^2)^2$