Let $m\in N$ with $m\geq 2$. Then for all $n\in N$: $m^n> n$. I am having a lot of trouble with this problem for some reason. I need to show that if $m\in \mathbb{N}$ with $m\geq 2$, then for all $n\in \mathbb{N},$ $m^n > n$. If I could get some pointers in the right direction, that would be great. Thank you. This is what I have so far:
Base Case: let $m=2$ and $n=1$. Then $2^1=2>1$. So the base case is true.
Induction Hypothesis: Assume that $k^l>l$ for some $k\in \mathbb{N}$ with $k\geq 2$ and $l\in \mathbb{N}$.
Induction step for $l$: $k^{l+1}=k\cdot k^l>kl=k\cdot(l+1)-k$. Since $k\geq 2$, We know that $k\cdot(l+1)-k>k(l+1)>l+1$. Thus, we have $k^{l+1}>l+1$. 
I am having trouble with the induction step for $k$. 

The base step should be for all $m$.
Base step:  $n= 1$.  If $m\ge 2$ then $m^n  = m^1 = m \ge 2 > 1=n$.  That's a base step for any $m$.
Induction step:  $n=k$.  If $m^k > k$ then
$m^{k+1} = m*m^k > m*k \ge 2k = k + k \ge k + 1$.
And that's it....
