# For every $L \subseteq \{a\}^*$ we have $L^*$ is regular [duplicate]

I want to prove that

for every language $L\subseteq \{a\}^*$, the laguage $L^*$ is regular.

every element of $\{a\}^*$ is equivalent to a natural number based on its length, and also every language $L$ is equivalent to a subset of natural number. after that base on lemma that says

Let $S$ be an infinite subset of $N$ and $d = gcd(S)$, there exists a finite subset $U$ of $S$ such that $d = gcd(U)$

I tried to prove the problem, first I claim that for every subset of natural number like $X$ that $d_x = gcd(X)$ there exist a number $p$ such that for every number $y$ greater than $p$ that is multiples of $d_x$ we can write

$y = \sum_{x \in X, t_x\geq0} t_x x$

so now we can build $DFA$ for $gcd(L)^*$ because $L$ contains a finite subset like $U$ such that $gcd(L) = gcd(U)$ so for every $w$ with length larger than $p$ languages $U^*$ and $L^*$ are behave the same and for words with length smaller than $w$ there is finite difference and also there is finite difference between $U^*$ and $gcd(U)^*$ because for every word with length greater than $p$ they also behave the same. so we can build $DFA$ for $L^*$ so it's Regular.

I think there are some holes in my proof. I am not sure about the claim and about the correctness.