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I do not understand the proof for this. I know that every word in $(A^*)^*$ is made up of words from $A^*$, and that this is made up from words in $A$. But how does this help with showing that $(A^*)^*$ is a subset of $A^*$.
Knowing that $A$ is regular is not relevant.
Hint. Just replace "made up of words of $A^*$" by "is a concatenation of words of $A^*$" in your sentence and you will have the answer.
A concatenation of words of $A^*$ gives a word which is in $A^*$, because of the very definition of $A^*$. This gives the result.
Moreover, one can prove that $^*$ is a closure operator, which includes this property.
Required, but never shown