Let ∑ = {a} and L1 = ∑*
Then we can say that L1 is regular.
Now take any arbitrary string x belongs to ∑*
then x = am for some integer m
Now, m must have a prime factor p
Hence m = p*k (for some integer k)
Hence, am = ap . ap . ap . ap .....(concatenating the string ap k times)
Now, ap belongs to L (given in the question ) which means (ap )k belongs to L*
Hence am belongs to L* which means x belongs to L*
So, for any arbitrary string x belongs to ∑* we get that x belongs to L*
Hence ∑* = L*
Thus L* is a regular language.