$L = \{ a^{j!} \mid j \geq1\}$ is not context free by pumping lemma
Suppose $L$ is context free. Then, by pumping lemma, there exist two integers $h \ge 1$ and $1 \le x \le \min\{h!, c_L\}$, where $c_L$ is a constant depending on $L$, that satisfy the following:


$a^{h!} \in L$
$a^{h!-x} a^{xn} \in L$ for every $n \ge 0$.1


Now you only need to show that, for some value of $n$, $a^{h!-x} a^{xn} \not\in L$.
Equivalently, you want to find a value of $n$ (possibly depending on $h$) such that
$h! + (n-1)x$ cannot be written as $j!$ for any integer $j \ge 1$.

For example, if $h = 1$ then $x=1$ so you can pick $n=0$ to obtain $h! + (n-1)x = 0$.

When $h \ge 2$, you can pick $n=2$ so that $h! + (n-1)x = h! + x$, which is not a factorial of an integer as the following inequalities show:
$$
h! &lt; h! + x \le 2 \cdot h! &lt; (h+1)!
$$


1. This is a consequence of the general form of the pumping lemma and of the fact that all the words in your language contain only the character a.