I'm trying to write a simple cryptography proof in LaTeX but am doing a horrible job. It's been a long time since I've written a proof successfully. Please help me complete my proof.

Definitions:

A function $f$ ($\mathbb{N} \to \mathbb{R}$) is negligible if for every positive integer $c$ there exists an $N$ such that for all integers $n \ge N$, it holds that $|f(n)| < \dfrac{1}{n^c}$.

I'm trying to prove that the function $\epsilon(n) = 2^{-n}$ is negligible.

Here's what I've got so far:

We want to show that $2^{-n}$ is smaller than $\dfrac{1}{n^c}$ for every integer $n$ greater than $N$, and that such an $N$ exists for every integer $c$.

We'll start with $\epsilon(n) = 2^{-n} < \dfrac{1}{n^c}$ and take the base-2 logarithm of both sides.

Simplifying, we have $-n < \log_2{n^{-c}}$.

Simplifying further, it becomes

$-n < -c \log_2{n}$

Multiplying by -1 on both sides, this is

$n > c \log_2{n}$

Now we just have to choose an $n$ for which this inequality always holds. We'll choose $n = 2^c$.

Substituting, we have

$2^c > c \log_2{2^c}$

Simplifying gives us

$2^c > c^2$

----------

I think I'm on the right track, but I'm not sure how to tie this part back into the rest of the proof. For instance, this doesn't hold for all $c$, e.g. for $c = 2$ or $c = 3$. This is supposed to be true for all positive integers $c$. But I think it works for $c = 5$ and up.  So there's something to do with the "for every integer $n$ greater than $N$" part that is causing me trouble. Maybe I have to add another polynomial to how I define $n$? I don't know. Please suggest how the proof can be completed and use LaTeX as I have been doing.
The proof can be completed as follows:

We want to show that $\epsilon(n) = 2^{-n}$ is smaller than $\dfrac{1}{n^c}$ for every integer $n$ greater than $N$, and that such an $N$ exists for every positive integer $c$.

We'll start with $\epsilon(n) = 2^{-n} < \dfrac{1}{n^c}$ and take the base-2 logarithm of both sides.

Simplifying, we have $-n < \log_2{n^{-c}}$.

Simplifying further, it becomes

$-n < -c \log_2{n}$

Multiplying by -1 on both sides, this is

$n > c \log_2{n}$

Next, we need to find an $N$ for which the inequality holds for all $n \ge N$. One way to do this is to find the maximum value of $c \log_2 n$ for $n \ge 2^{c}$. We have $c \log_2 n \le c \log_2 2^{c} = c \cdot c$. Hence, we can set $N = \max(2^{c}, c^2)$. Then, for all $n \ge N$, it holds that $n > c \log_2 n$ and hence $2^{-n} < \dfrac{1}{n^c}$.

In conclusion, $\epsilon(n) = 2^{-n}$ is a negligible function according to the definition, since for every positive integer $c$ there exists an $N$ such that for all integers $n \ge N$, it holds that $2^{-n} < \dfrac{1}{n^c}$.