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 ... \} $
Is star closure of reverse of language equivalent to reverse of closure of that language
formal languages;automata
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 ${(L^*)}^R={(L^R)}^*$