What is the error in this proof that the sum of the $n$-th roots of unity is $0$? When I was in high school, in one of my exams, I was asked to prove that sum of the $n$-th roots of unity is $0$. Instead of the usual proof that was taught in class, I answered with a proof that struck me in the exam hall. If $1, \alpha, \alpha^2, \alpha^3...\alpha^{n-1}$ are the roots :
$$
1+ \alpha+\alpha^2+ \alpha^3...+\alpha^{n-1}=
\\
\alpha^n+ \alpha+\alpha^2+ \alpha^3...+\alpha^{n-1}=
\\
\alpha(1+ \alpha+\alpha^2+ \alpha^3...+\alpha^{n-1})
$$
Which is possible only if this sum is $0$ as $\alpha$ is not equal to $1$. 
My teacher marked the proof wrong saying that it was inadequate. What's the problem with it ?
 A: There is absolutely nothing wrong with it. Perhaps your teacher would have understood better if you typed some intermediate step, e.g. "$\alpha + \alpha^2 + \alpha^3 + \ldots + \alpha^{n-1} + \alpha^n = $" between the second and third line and or had said more explicitly that this simplifies to $S = \alpha S$ with $S$ the sum we want to compute before drawing the conclusion. But these are just clarifications of things that are already there. I really think your proof is correct!
A: I would make it a bit more explicit. From your equation
$$
(1+\dotsb+\alpha^{n-1}) = \alpha (1+\dotsb+\alpha^{n-1}) \iff \\
(1-\alpha)(1+\dotsb+\alpha^{n-1}) =  0 \iff \\
(\alpha = 1) \vee (1+\dotsb+\alpha^{n-1} = 0)
$$
one can derive the logical equivalent statement thet either $\alpha = 1$ (which you said it is not) or the sum must vanish.
Sometimes the $\cdots$ notation is criticized, but in this case it is precise enought IMHO, and could be easily improvied by using $\sum$ notation.
A: There is a small chance that the teacher misread your second equality because you changed the order at the same time that you factored out $\alpha$.
$$\begin{align}S&:=1+ \alpha+\alpha^2+ \alpha^3...+\alpha^{n-1}\\
&=\alpha^n+ \alpha+\alpha^2+ \alpha^3...+\alpha^{n-1}\\
&=\alpha(\alpha^{n-1}+1+ \alpha+\alpha^2+...+\alpha^{n-2})\\
&=\alpha S\end{align}
$$
might be slightly more convincing.
