Proving that $\inf\{ n^{1/n} > 1 : n \in \mathbb{N} \}=1$ using only that $1 + ma \le (1+a)^m$ for every $a>0$ and $m \in \mathbb{N}$

Question

I'm trying to find the infimum of the following:

$S = \{ n^{1/n} > 1 : n \in \mathbb{N} \}$

using the fact that $1 + ma \le (1+a)^m$ where $a>0$, $m \in \mathbb{N}$

What I tried:

I know $\xi: \mathbb{N}-\{1,2\} \mapsto \mathbb{R}^+$, $\xi(n)=n^{1/n}$ is decreasing, and already know $\inf S=1$.

But using the fact above, I would like to show the following.

$$ \forall \epsilon >0, \exists n>2 : (1+\epsilon)^n \ge n $$

But I can't proceed.