Perhaps if we see it written in mathematics it will be clearer. Let us list down what we know:
$(1)\;\;\;\;\;\;\;\forall\,n\in\Bbb N\,\,,\,2^n>n$
$(2)\;\;\;\;\;\;\;\;\forall\,\,\epsilon>0\,\,\,,\,\,\,\frac{1}{2^n}<\epsilon\Longleftrightarrow 2^n>\frac{1}{\epsilon}$
$(3)\;\;\;\;\;\;\;\text{Thus, if}\,\,n>\frac{1}{\epsilon}\,\,,\,\text{ we get:}$
$2^n\stackrel{\text{by}\,\,(1)}>n\stackrel{\text{if (3)}}>\frac{1}{\epsilon}\stackrel{\text{by (2)}}\Longrightarrow 2^n>\frac{1}{\epsilon}\Longleftrightarrow \frac{1}{2^n}<\epsilon$
and then we're done.
Note that in the right arrow $\,\Longrightarrow\,$ in the last line above, we used transitivity of the relation $\,x>y\,$:
$x>y>z\Longrightarrow x>z$
Finally, how do we prove that there exists some $\,n\in\Bbb N\,$ that fulfills (3) above, no matter what $\,\epsilon\,$ , and thus what $\,1/\epsilon\,$ , is?
Very simple: this is the Archimedean Property of the natural numbers, and you can read about it here