L ={ $a^{m^n}$ | $m$>$n$ } is Regular or not by pumping Lemma
Let $A = \{ m^n : m &gt; n \}$, so that $L = \{a^k : k \in A\}$. It is not clear whether $m,n \geq 0$ or $m,n &gt; 0$. In both cases, every $m &gt; 2$ is in $A$ and $0 \notin A$. If $m,n &gt; 0$ then $1 \notin A$, while if $m,n \geq 0$ then $1 \in A$. In the former case, $L = aa^+$, and in the latter, $L = a^+$. In both cases $L$ is actually regular.

We can also consider the other interpretation, $A = \{mn : m &gt; n\}$; this interpretation is wrong, but it can shed light on your proof. Again we find that every $m &gt; 2$ is in $A$, and that $1 \notin A$. If $m,n$ are allowed to be zero then $0 \in A$, and otherwise $0 \notin A$. In the former case $L = aa^+ + \epsilon$, and in the latter, $L = aa^+$. In both cases $L$ is regular. So something must have been wrong with your proof.