So I want to prove that for natural numbers $m,n,k$, if $k$ is no $0$ or $1$, and if $m, then $k^m. I jotted down the following proof, but I'm not sure how acceptable it is. I only have use of basic set theoretic properties of the naturals with addition, multiplication, and powers. I also have use of cancellation.
Let $ K=\{k\in\omega\ |\ k^m
Since $m there exists a $p^+\in\omega$ such that $m+p^+=n$. Consider $2$. I claim $2^{p^+}>1$. Now $p^+\neq 0$, since $0$ is not in the range of the successor. If $p^+=1$, then from the work above, $2^1=2>1$. Suppose then that $2^{p^+}>1$. Then $2^{p^{++}}=2^{p^+}\cdot 2>1$. This follows by transitivity since $1<2^{p^+}$, and $1<2$ implies $2^{p^+}\cdot 1<2^{p^+}\cdot 2$, that is $2^{p^+}<2^{p^{++}}$. Then from $1<2^{p^+}$ we have $2^m=2^m\cdot 1<2^m\cdot 2^{p^+}=2^{m+p^+}=2^n$, so $2\in K$. By the same reasoning we can show that any $k\neq 0,1$ is in $K$. If $k\neq 0,1$, then $1. Again, I claim $k^{p^+}>1$. If $p^+=1$, we are done. So suppose $k^{p^+}>1$. Then since $1, we have $k^{p^+}=k^{p^+}\cdot 1, so $1 by transitivity, and so $k^{p^+}>1$ for all $p$, since $p^+\neq 0$. Then from $1 we have $k^m\cdot 1, so $k\in K$, and the result follows.
All in all, this approach seems very ugly. Is there a way to clean it up? I wanted to use the time tested method of induction, but assuming $k^m seems to have little to do with showing $(k^+)^m<(k^+)^n$, since I don't know how to distribute powers. Thanks for any constructive criticism.