Is star closure of reverse of language equivalent to reverse of closure of that language is the following true 
$ (L^R)^* = (L^*)^R $ 

I tried the following to prove it true.
let u,v belong to L then $ L^* = \{ u,v, uu, vv, uv, vu ... \} $
and $ (L^*)^R = \{ u^R, v^R, u^Ru^R, v^Rv^R, v^Ru^R, u^Rv^R ... \} $

now $ L^R = \{ u^R, v^R \} $ so 
$(L^R)^* = \{ u^R, v^R, u^Ru^R, v^Rv^R, u^Rv^R, v^Ru^R ... \} $
Take any $w\in {(L^*)}^R$. Then $w$ can be writen as $w=(u_1\cdot u_2\cdots u_n)^R$, with $u_i\in L$. We have

$$
w =(u_1\cdot u_2\cdots u_n)^R =u_n^R \cdot u_{n-1}^R \cdots u_1^R,
$$

and therefore $w\in (L^R)^*$.

Assume now that  $w\in {(L^R)}^*$, then by the same argument 
$$
w =u_1^R \cdot u_{2}^R \cdots u_n^R= (u_n\cdot u_{n-1}\cdots u_1)^R ,
$$
and hence $w\in {(L^*)}^R$.

As a consequence ${(L^*)}^R={(L^R)}^*$