Sum of roots of unity using a particular argument On this post it claims that for $n>1$ the sum of the $n$th roots of unity may be written as $S=1+\omega_1+\omega_2+\cdots+\omega_{n-1}$ and then multiplied on both sides by $\omega_1$, where $\omega_1\neq1$, to obtain $S=\omega_1 S$. Hence, $S=0$. 
Question: How is $S=\omega_1S$ obtained by multiplying both sides by $\omega_1$? That is not clear to me at all (that really seems to be the main argument). Everything else is fine, but I do not see how $S=\omega_1 S$ immediately follows from multiplying both sides by $\omega_1$. Any insight would be appreciated. 
 A: We have:
$1\cdot\omega_1=\omega_1$
$\omega_1\cdot\omega_1=\omega_2$
$\omega_2\cdot\omega_1=\omega_3$
etc.
so multiplying the original equation by S gives
$\omega_1S=\omega_1+\omega_2+\dots +\omega_{n-1}+\omega_n=\omega_1+\omega_2+\dots +\omega_{n-1}+1=S$
and the result follows.
(Note that this works because of the unit complex circle. Relative to the x-axis (the real axis), $\omega_1$ has an argument of $\frac{2\pi i}{n}$, and $\omega_k$ has an argument of $\frac{2k\pi i}{n}.$)
A: As the product of two roots of unity is a root of unity, multiplication by $\omega_1$ of all elements in the list $(1, \omega_1,\dots ,\omega_n)$ results in a permutation of these elements. Hence, as their sum is a symmetric function of them, it is invariant by a permutation  of its terms.
Note:
Another proof consists in noting the $n$-th roots of unity are the roots of the the polynomial $X^n-1$.  $S=0\;$ simply results from Vieta's relations.
A: 
for $n>1$ the sum of the $n$th roots of unity may be written as $S=1+\omega_1+\omega_2+\cdots+\omega_{n-1}$ and then multiplied on both sides by $\omega_1$, where $\omega_1\neq1$, to obtain $S=\omega_1 S$. Hence, $S=0$. 

The implied notation is $\omega_k=e^{i \frac{2k \pi}{n}}=\omega_1^k\,$ for $\,k=0,\cdots,n-1\,$.
Then $S=\omega_1^0+\omega_1^1+\cdots+\omega_1^{n-1}\,$, so $\omega_1 \cdot S = \omega_1^1 + \omega_1^2+\cdots+\omega_1^n=S\,$ since $\omega_1^n=1=\omega_1^0\,$.

As far as the proposition itself, the direct proof that $\,S=0\,$ is using Vieta's relations, since $\omega_k$ are by definition the roots of $z^n-1=0\,$.
