There is a relation $R$ and $S$ on set $U$. Given $R$ is transitive.
I need to prove $(R\circ S \circ R)^n \subseteq (R \circ S)^n \circ R$ for all $n \geq 1 $
My Attempt:- Tried doing this by induction
for base case $n=1$ its trivial as $(R\circ S \circ R)^1 \subseteq (R \circ S)^1 \circ R = (R\circ S \circ R) \subseteq (R \circ S) \circ R$
Took Induction Step as : $(R\circ S \circ R)^n \subseteq (R \circ S)^n \circ R$
now for $n=n+1$
$(R\circ S \circ R)^{n+1} $
$R\circ S \circ R \circ (R\circ S \circ R)^n $ $\qquad$ { Def. of exponents }
$\subseteq R\circ S \circ R \circ (R\circ S)^n \circ R$ $\qquad$ { Using $I.H$ and monotonicity }
I am stuck here and do not know how to proceed further. Can anybody try to help me by providing a hint.
HINT: Expand $(R\circ S\circ R)^{n+1}$ the other way,
$$(R\circ S\circ R)^{n+1}=(R\circ S\circ R)^n\circ R\circ S\circ R\subseteq(R\circ S)^n\circ R\circ R\circ S\circ R\;,$$
and apply the transitivity of $R$ to $R\circ R$.