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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., $$ (S^2/4)n \frac{1}{\tan(\pi/n)}. $$
Hint: "Triangulate" your polygon, i.e. decompose it in $n$ triangles and measure the area of one of them (they are all same). Note that they are all isosceles (because the polygon is regular) and you can easily compute their angles. So you may take the altitude of it passing through the center of the polygon and then apply basic trigonometric formulas.