L ={ $a^{m^n}$ | $m$>$n$ } is Regular or not by pumping Lemma L ={ $a^{m^n}$ | $m$>$n$ }
I am bit confuse whether to consider this language as L = $(a^{m})^{n}$ OR  L = $a^{\left(m^n\right)}$.        

If it is considered as L = $(a^m)^{n}$ then want to check it is Regular or not by Pumping Lemma.
I tried by following way.    

$m$ > $n$ so we can take $m's$ value as $n+1$.
So $L$  = $(a^{n+1})^{n}$  

Let n be the Pumping Lemma Constant.
Then by pumping Lemma $u$$v^{i}$$w$ in $L$ for every $i$ $&gt;=$ $0$.  

$1&lt;= |v| &lt;= n$   

$uv &lt;= n$.  

So here $u$ = $(a^{n+1})^{n-1}$      

$v$ = $(a^{n+1})^{i}$  and    

w = $\epsilon$   

By taking $i = 2$ ,
I am getting string as $(a^{n+1})^{n+1}$Which is not possible. Because it means $m=n$.
So above string is not Regular. 

But different solutions giving different answers. So confused whether it is Regular or not. And is there any problem in my solution ?
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.