Derivation of the formula for the area of a regular polygon given the side length. There are many formulas for finding the area of a regular polygon- but this is the one I am interested in:
$$A=\frac{S^2n}{4\tan(\frac{\pi}{n})}$$
where $n$ is the number of sides, and $S$ is the length of each side. Like I said, I would like to know how it is derived.

Well, it's $n$ times the area of a single "wedge" of the polygon. So I just need to show the wedge has area $S^2/(4 \tan (\pi/n))$. 
For that, consider a circle of radius $R$, with a wedge going from $-\pi/n$ to $\pi/n$. The area of that wedge is twice the area of the half-wedge, from $0$ to $\pi/n$. That area in turn is half the base times the height, i.e., it's $\frac{1}{2} R \sin (\pi/n) R \cos(\pi/n)$. So one wedge has area 
\begin{align}
W &= R \sin (\pi/n) R\cos(\pi/n)\\
& = R^2 \sin^2 (\pi/n) \frac{\cos(\pi/n)}{\sin(\pi/n)}\\
\end{align}
Now $R \sin(\pi/n)$ is half the length of a side of the polygon, i.e. it's $S/2$, so this becomes
\begin{align}
W &= R \sin (\pi/n) R\cos(\pi/n)\\
& = R^2 \sin^2 (\pi/n) \frac{\cos(\pi/n)}{\sin(\pi/n)}\\
&= (S/2)^2 \frac{\cos(\pi/n)}{\sin(\pi/n)}\\
&= (S^2/4) \frac{1}{\tan(\pi/n)}.
\end{align}
Thus the total area is $n$ times this, i.e., 
$$
(S^2/4)n \frac{1}{\tan(\pi/n)}.
$$
