Prove that Archimedean Property implies that $\displaystyle \lim_{n\to \infty} \frac{1}{n}=0$ Problem:

Prove that Archimedean Property implies that $\displaystyle \lim_{n\to \infty} \frac{1}{n}=0.$

I am very curious how to prove this.  To start off, we assume the Archimedean Property, or there exists an $\epsilon>0$ such that, $\dfrac{1}{n} < \epsilon$, for any natural number $n$.  But, from there I am simply lost.  Can someone help me out here?
 A: What does it mean for $a_n$ to satisfies $lim_{ n\to \infty} a_n = 0$?
It means that for every $\epsilon>0$ there is a $N$ such that, if $m>N$ then
$|a_m - 0| < \epsilon$
Back to our case:
Here $a_n = \frac{1}{n}$, then
$lim_{ n\to \infty} \frac{1}{n} = 0$ reads:
For every $\epsilon >0$ there is a $N$ such that, if $m>N$ then
$| \frac{1}{m}| = \frac{1}{m} < \epsilon$.
Can you finish from here?
A: Assume that $\displaystyle \lim_{n \to \infty}\frac{1}{n} \not= 0$, then by definition of limits, we have have: 

$\exists \epsilon>0$ such that $\forall N \in \mathbb{N}$, $\exists n \geq N$ for which $\displaystyle \left| \frac{1}{n}-0 \right| > \epsilon$

Since $n \geq N$ implies $1/n \leq 1/N$, the above actually implies that $\forall N \in \mathbb{N}$, $\displaystyle \frac{1}{N} \geq \left| \frac{1}{n}-0 \right| > \epsilon$, which in turn implies that $\forall N \in \mathbb{N}$, $\displaystyle N<\epsilon$, which contradicts the Archimedean property of real numbers since $\epsilon$ is a real number larger than any positive integer.
