Proving $({L_A}^* L_B)^+ = (L_A \cup L_B)^*L_B$ I'm trying to prove the following identity:
$$ ({L_A}^*\  L_B)^+ = (L_A \cup L_B)^*L_B $$

which is clearly true as both sides will exactly match "any sequence of A and B ending with B" :

$$ B, AB,AAB, ABBABAB, ... $$

I tried thinking about the set properties and even tried to reasoning about the NFA constructed by the the two sides of the equation but to no avail so far.

Does anyone know how to prove it?
Although I can't do the proof completely right now, maybe looking at $(L_A^* L_B)^+ \subseteq (L_A \cup L_B)^*L_B$ and $(L_A \cup L_B)^*L_B \subseteq (L_A^* L_B)^+$ helps here. So from left to right: If you take a word $w \in (L_A^* L_B)^+$, can you split it up into $0..m$ subwords of the form $(L_A \cup L_B)$ and one postfix-subword of form $L_B$? I hope this helps you.