[Solved]: 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 ${w in a^* mid |a| > (x-1)(y-1)-1}$. 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 ${w in (a^m)^* mid |w| > m^2[(x/m – 1)(y/m – 1) – 1]}$, 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 $a^{10}$.