Kleene star of an infinite unary language always yields a regular language
Suppose there are two words in the language whose lengths are relatively prime. Let these lengths be $x$ and $y$. We know (see this) that by adding these numbers to each other repeatedly, we can get any number greater than $(x - 1)(y - 1) - 1$. So if $x$ and $y$ are $13$ and $7$, we can write any number greater than $72$ as a linear combination of $7$ and $13$. What this means for us: $L_2^*$ consists of some arbitrary finite language (regular, as all finite languages), in union with the language $\{w \in a^* \mid |a| &gt; (x-1)(y-1)-1\}$. This language is regular since it is the language of all words with a finite set of words removed. Since $L_2^*$ is a union of regular languages, it must also be regular.

If all words in $L_2^*$ have lengths which share a greatest common factor (call this common factor $m$), then repeat the above argument, but instead of using string lengths, use string lengths divided by $m$. In this case, $L_2^*$ will be the concatenation of an arbitrary finite language (regular) and the language $\{w \in (a^m)^* \mid |w| &gt; m^2[(x/m - 1)(y/m - 1) - 1]\}$, also regular (since $(a^m)^* is regular and we are removing finitely many words from it).

For example, suppose all words in $L$ have a GCF of 2, and the language contains the words $a^4$ and $a^{10}$. We have $m = 2$, $x/m = 4/2 = 2$, and $y/m = 10/2 = 5$, which are relatively prime. Therefore, we know that we can get any word whose length is multiple of $m$ if the length is greater than $m^2[(x/m - 1)(y/m - 1) - 1] = 2^2[(2 - 1)(5 - 1) - 1] = (4)(3) = 12$ by concatenating $a^4$ and $a^{10}$.