Proving that $L = \{ a^{n!} \ | \ n \geq 0 \}$ is not regular
The easiest way is probably to use the pumping lemma on the complement language $\overline{L} = \{ a^m : m \neq n! \}$. Suppose that $L$ were regular. Than $\overline{L}$ would be regular. According to the pumping lemma, there is a pumping length $p$ such that every word $w$ of length at least $p$ in $\overline{L}$ has a decomposition $w = xyz$ such that $|xy| \leq p$, $y \neq \epsilon$, and $xy^iz \in \overline{L}$ for all $i \geq 0$. Take $w = a^{p\cdot p!} \in \overline{L}$ (without loss of generality, $p &gt; 1$). Suppose that the decomposition is $w=xyz$, where $y=a^t$. Since $t \leq p$, it follows that $t \mid p!$. But then $xy^{1+p!/t}z = a^{p\cdot p! + p!} = a^{(p+1)!} \notin \overline{L}$.