Proof involving $R^n$ and the transitivity of a relation

Question

I want to prove:

R is the relation on the set A. If R is transitive, then $R^n$⊆$R^{n-1}$ for n = 2, 3, 4,...

I'm having trouble approaching this proof, I've started my proof by induction in which my base case is $R^2$

I know $R^2$ = R ○ R by the recursive definition of the powers of R and I know that R is transitive, thus
$(a,b) ∈ R ∧ (b,c) ∈ R \implies (a,c) ∈ R$

However, I'm having trouble proceeding from here. Can anyone help?

Answer 1

If $(x,y) \in R^2$ then there exist a $z \in A$ such that $(x,z) \in R$ and $(z, y) \in R$, by the definition of the composition of relations. Now transitivity implies that $(x,y) \in R$. Therefore $R^2 \subset R$. Now use the strong induction to prove the general case.