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By pumping lemma, we choose the word $w=a^{p^2}$ that the decomposing is $[a^sa^ta^{p-s-t}]^p$ such that $u=a^s,v^i=a^t,x=a^{p-s-t}$
$[a^sa^{it}a^{p-s-t}]^p=[a^{p+it-t}]^p$
We choose i=p+1,we get $ [a^{(1+t)p}]^p $ $=a^{(1+t)p^{2}}$
we see that the power $(1+t)p^{2}$ has 4 dividers thus doesn't belong to the Language L where the power of a is $p^2$ with 3 dividers.
Is my answer correct?
Instead of looking at primes I would prove this:
The language { a^k: k is an element of S } where S is an infinite set of non-negative integers with arbitrary large gaps between elements is not regular.
It's not more difficult than the statement you have, and it proves immediately that languages like $a^{n^2}$, $a^{n!}$ etc. are all non-regular.
