So, let's finish this. Starting from Qiaochu's formula,
$$\prod_{k=0}^{n-1} (r^2 \cos^2 (2 \pi k/n) + s^2 \sin^2 (2 \pi k)/n))=1$$.
Each factor is
$$\left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) + i s \sin(2 \pi k/n) \right) \left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) - i s \sin(2 \pi k/n) \right) =$$ $$\left( \frac{r+s}{2} e^{2 \pi i k/n} + \frac{r-s}{2} e^{-2 \pi i k/n} \right) \left( \frac{r+s}{2} e^{2 \pi i k/n} - \frac{r-s}{2} e^{-2 \pi i k/n} \right).$$
Putting $u=(r+s)/2$, $v=(r-s)/2$ and $\zeta=e^{2 \pi k/n}$, we have $$\prod_{k=0}^{n=1} (u + v \zeta^{2k}) (u - v \zeta^{2k})$$ If $n$ is odd, this is $(u^n+v^n)(u^n - v^n)= (u^{2n} - v^{2n})$. If $n$ is even, I get $(u^{n/2} - v^{n/2})^2 (u^{n/2} + v^{n/2})^2 = (u^n-v^n)^2$. So either $u^{2n} - v^{2n}=1$ or $u^n-v^n=1$ or $y^n-v^n=-1$. These are all forms of FLT.