# Simplying this regular expression

I am trying to prove (without converting it into Automata, through just simplification) that the following two regular expressions are equal:

$(\epsilon + 0^*1^+0)^* = \epsilon + (0+1)^*10$

here is what I have got so far:

$$(\epsilon + 0^*1^+0)^* = \\ (0^*1^+0)^* = \\ (0^*1^*10)^* = \\ \epsilon + (0^*1^*10)^+ = \\ \epsilon + (0^*1^*10)(0^*1^*10)^* = \\ \epsilon + 0^*1^*10(0^*1^*10)^*$$

I seem to be stuck, I am not sure if I am doing it the correct way, or if there is a better way to prove this.

I know that $(a+b)^* = (a^*b^*)^*$ is true and I think this will certainly come in handy for the problem above but I am not sure how to prove $(a+b)^* = (a^*b^*)^*$ either.

The regular expression on the right consists of the empty word along with all words ending in $10$. All words accepted by the regular expression on the left are of this form. It remains to show that all words accepted by the regular expression on the right are also accepted by the one on the left. This is clear for the empty word, so it remains to show the following:
Every word ending in $10$ is accepted by $(0^*1^+0)^*$.
Indeed, every word ending in $10$ has the form $$0^{a_1} 1^{b_1} 0^{a_2} 1^{b_2} \cdots 0^{a_n} 1^{b_n} 0,$$ where $b_1,a_2,b_2,\ldots,a_n,b_n \geq 1$. We can decompose this word as $$(0^{a_1} 1^{b_1} 0) (0^{a_2-1} 1^{b_2} 0) (0^{a_3-1} 1^{b_3} 0) \cdots (0^{a_{n-1}-1} 1^{b_n} 0) \in (0^*1^+0)^*.$$