How to prove that $R_1(R_2R_1)^* = (R_1R_2)^*R_1$; Theory of computation I need a formal proof for this property (or hints):
Let $R_1$ and $R_2$ be to regular expressions : 
$R_1(R_2 R_1)^* = (R_1 R_2)^*R_1$
I don't know if the way I'm solving the problem is totally fine, but I'll share it : 
Let $Q_1=R_1(R_2R_1)^*$ and $Q_2=(R_1R_2)^*R_1$ ; there are two cases : 


*

*case 1 : $(R_2R_1)^*=\epsilon$ $\implies$  $Q_2=R_1$ ; we can find $R_1$ in the language described by $Q_2$ if $(R_1R_2)^* = \epsilon$ ; [...] in this case : $(R_2R_1)^*=\epsilon \Leftrightarrow (R_1R_2)^* =\epsilon $ and then $Q_2=Q_1$ .

*case 2 : $(R_2R_1)^*\neq \epsilon$ ( then $(R_1R_2)^*\neq\epsilon$ ) $\implies Q_1=R_1(R_2R_1)...(R_2R_1) = (R_1R_2)(R_1R_2)...(R_1R_2)R_1 = Q_2$ 
( $\epsilon$ = empty string )
 A: Problem
$(R_2R_1)^*\neq(R_2R_1)...(R_2R_1)$ since $(R_2R_1)...(R_2R_1)$ has an "end" to it, and you haven't stated whether or not $\epsilon\in R_2R_1$ (which it is in $(R_2R_1)^*$ by definition). 
Suggestion
You have the right idea, your proof just has a few technical problems. Since $R_1(R_2R_1)^*$ and $(R_1R_2)^*R_1$ are sets, you need to prove that,  $R_1(R_2R_1)^*\subseteq (R_1R_2)^*R_1$ and $R_1(R_2R_1)^*\supseteq (R_1R_2)^*R_1$ to show that $R_1(R_2R_1)^*= (R_1R_2)^*R_1$ . To do this you should first prove that if $x\in (R_1R_2)^*R_1$ and $x\neq \epsilon \implies x\in (\prod\limits_{i=1}^{n}R_1R_2)R_1$ for some $n$. With that you can use your trick to show that $x\in R_1(R_2R_1)^*$. 
Notes
$\prod\limits_{i=1}^{n} R_2R_1=(R_2R_1)...(R_2R_1)$ where there are $n$, $R_2R_1$'s.
Example: $\prod\limits_{i=1}^{3} R_2R_1=(R_2R_1)(R_2R_1)(R_2R_1)$
A: It suffices to come back to the definition of the star operator:
$$
L^* = \bigcup_{n \geqslant 0} L^n
$$
Now, using the fact that $R_1(R_2R_1)^n = (R_1R_2)^nR_1$, one gets
$$
R_1(R_2R_1)^* = R_1\Bigl (\bigcup_{n \geqslant 0}\ (R_2R_1)^n \Bigr) = \bigcup_{n \geqslant 0}\ R_1(R_2R_1)^n = \bigcup_{n \geqslant 0}\ (R_1R_2)^nR_1 = \Bigl (\bigcup_{n \geqslant 0}\ (R_1R_2)^n \Bigr) R_1 = (R_1R_2)^*R_1
$$
