Let $m\in N$ with $m\geq 2$. Then for all $n\in N$: $m^n> n$.

Question

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 $k$.

Answer 1

Since $m^n \ge 2^n$. we just have to prove that $2^n > n$.

Suppose we know that $2^k > k$, then we have

$$2^{k+1}>2k=k+k \ge k+1$$

Answer 2

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....

Answer 3

Base case:

$$m>1\implies m^1>1$$ so the claim holds for $n=1$.

Inductive step:

$$m>1,m^n>n\implies m^{n+1}=m\cdot m^n>mn\ge n+1$$

so if the claim holds for $n$, it holds for $n+1$.