Proving that $L = \{ a^{n!} \ | \ n \geq 0 \}$ is not regular Let $L$ a language over $X = \{a\}$ defined as follow : 

$$L = \{ a^{n!} \ | \ n \geq 0 \}$$

I want to prove that $L$ isn't regular, I have searched in the forum for an equivalent question, but I found nothing. If it's a duplicate, I apologize and I'll be glad if you provide me the link of the duplicated question.

I'll show here what I have done, using the pumping lemma.



Suppose that $L$ is regular, let $n \in \mathbb{N}^*$ and $\omega = a^{n!}$. We have $\omega \in L$ and $|\omega| \geq n$. Thus, there exist $x, y, z \in X$ so as :


$a^{n!} = xyz$
$y \neq \epsilon$
$|xy| \leq n$
$xy^kz \in L, \quad \forall k \geq 0$


Because of fact 3, $xy = a^k$ and $y = a^j$ with $k \leq n$ and $1 \leq j \leq n$. Thus :

$$a^{n!} = a^{k - j}.a^{j}.a^{n! - k}$$

By fact 4, we must have $a^{k - j}.a^{n! - k} = a^{n! - j} \in L$. 

In addition, because $1 \leq j \leq n$, the following is true : 

$$n! - n \leq n! - j \lneq n!$$

The final result that I need is that $(n - 1)! \lneq n! - n$. Using this I can deduce :

$$(n - 1)! \lneq n! - j \lneq n!$$

Showing that $n! - j$ can't be the factorial of an integer, giving the contradiction $a^{n! - j} \notin L$. I have tried to prove the final result, but I didn't succeed. It is not true for the first values of $n$, I have tried to consider multiple cases but went for nothing.

So I ask for your advice, maybe I am going wrong and didn't applied the pumping lemma with a word that leads to contradiction.
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 $xy^{1+p!/t}z = a^{p\cdot p! + p!} = a^{(p+1)!} \notin \overline{L}$.