As stated in the Wikipedia article on roots of unity, the reciprocal of an $n$-th root of unity is its complex conjugate. They provide the following proof of this statement:
Let $z\in\mathbb{C}$ be a $n$-th root of unity $\Rightarrow$ $$\frac{1}{z}=z^{-1}=z^nz^{-1}=z^{n-1}=\overline{z}$$ I'm unable to figure out why the last equal sign holds.
On
You should draw a picture, I think there is no way around that.
A suitable picture will be $S^1$ in the complex plane, each $n$-th root marked on it with a dot. $z^n=1$ is $(1,0)$ on the $x$-axis. From this point there are two equidistant points, $z$ and $z^{n-1}$, to its left and right on the circle. Since complex conjugation is the same as mirroring the complex number around the $x$-axis it should be clear that $\overline{z}=z^{n-1}$.
It is always true that $z\overline z = |z|^2$, so when $|z| = 1$, we may conclude that $\overline z = 1/z$.
edit: and you know that roots of unity have $|z| = 1$ because $|z^k| = |z|^k$ (which you can prove by showing $|wz| = |w||z|$ and repeating) and hence any $n^\mathrm{th}$ root of unity has $|z|^n = 1$, which (since $|z|$ is a positive real number) means $|z| = 1$.