Proving language is not regular using pumping lemma

Question

Show that the language

                    $L =$$\left \{ a^{n!} : n\geq 1 \right \}$ is not regular using pumping lemma

My solution is :

Suppose L is regular

There exist some pumping length for L, let it be "m"

We don't know the value of m but whatever it is we can always choose n = m.

Then from the pumping lemma there exists $x, y, z \in Σ^{∗}$ such that w = xyz , |xy| ≤ m and |y| ≥ 1.

$w_{i} = xy^{i}z$

 y must contain entirely of a's

suppose $|y|=k$

the string obtained by i = 1 is,

w = $a^{m!+k}$

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Now after this i'm stuck because how can i prove $m!+k$ cannot be accepted by the language

also if i chose i = 0 then also i'm stuck

Answer 1

Pick $i$ such that $m! + ik \neq h!$ for some $h$, and you are done. In particular, if $i=1$, then $m! + k \leq m! + m \neq (m+1)!$.

Your proof otherwise looks good!