Problem as follows:
Let $R$ be a transitive relation. Let $aR^nb$ for $n \geqslant 1$, mean that there is a sequence of tuples: $$ ⟨a_0,a_1⟩,⟨a_1,a_2⟩, \ldots , ⟨a_{n-1},a_n⟩ $$ from $R$ such that $a_0 = a$ and $a_n = b$. Prove the following: "If $a R^n b$, then $a R b$" for all natural numbers $n \geqslant 1$, by mathematical induction.
Now, I understand why this assertion would be true, it's just proving it with mathematical induction that's giving me problems. (Check for $n = 1$, assume $n = k$, prove $n = k+1$ etc.)
Suppose that $a\mathop{R^n}b\implies a\mathop Rb$. Then, if $a\mathop{R^{n+1}}b$, there is a $c$ such that $a\mathop Rc$ and that $c\mathop{R^n}b$. But $c\mathop{R^n}b\implies c\mathop Rb$. So, $a\mathop Rc$ and $c\mathop Rb$. Since $R$ is transitive, $a\mathop Rb$.