limits - $limlimits_{nto infty}frac{n^k}{2^n}$ without l'Hopital

Let $\varepsilon >0$. Let $N>?$. For any integer $n>N$ we have

$$\frac{n^k}{2^n}<\varepsilon.$$
I don't know how to proceed here sensically.

I'd say we have to start with

$$<\frac{n^k}{2^N}$$

But what do we do here about the $n^k$?

Remember, I don't want to use l'Hopital or for me unproven limit laws like "exponents grow faster than powers"

Also, I don't want to use hidden l'Hopital, i.e. argumenting with derivatives. Since we don't even have proven derivative laws.

Answer

Let's take the logarithm. One gets

$$\log{n^k\over 2^n}=k\log{n}-n\log{2}=-n\left(\log{2}-k{\log{n}\over n}\right)$$

Now when $n\to\infty$ one has $\log{n}/n\to 0$ and so

$$\lim_{n\to\infty}\log{n^k\over 2^n}=-\infty$$

And so

$$\lim_{n\to\infty}{n^k\over 2^n}=0$$