I am studying formal language theory and have been asked to prove the following:
$\forall L, L^*=(L^*)^*$
I've started with
$def. L^* = \bigcup_{i \in \mathbb{N}} L_i, L_0=\{{\epsilon}\}, L_1=\{L\}, L_{i+1}=\{uv|u\in L_i, v\in L\}$
$then (L^*)^* = \bigcup_{i \in \mathbb{N}} L_i^*, L_0^*=\{{\epsilon}\}, L_1^*=\{L^*\}, L_{i+1}^*=\{uv|u\in L_i^*, v\in L^*\}$
$L^*L_0^*\in(L^*)^* = L^*\epsilon^* = L^*$ $\therefore L^* \subseteq (L^*)^*$
$L^*L_0^* = L^*, L^*L^*_1 = L^* (def), L^*L_{i+1}^*=L^*_iL^*=L^*L^*$ but the * operator is closed under concatenation, thus, $L^*L^* \in L^* \therefore (L^*)^* \subseteq L^*$
$\therefore L^* = (L^*)^*$
I simply don't have an intuition on whether this attempt at a proof is correct and I would appreciate any insight about it since the subsequent problems are to show $L^+=(L^+)^+$ and then to argue whether $(L^*)^+=(L^+)^*$