I came across an equation in a note but it did not compile in my mind. Here is the equation:
$$\prod_{k=1}^p(1+\varepsilon^kx)=1+x^p$$
where $\varepsilon=e^{2\pi i/p}$ and $p$ is an odd prime number.
I have tried to prove that every coefficient of $x^k$ equals $0$($1<k<p$) but can someone show me a more elegant proof?
The condition should be $p$ is an odd integer.
Let $f(x) = \prod_{k=1}^p(1+\epsilon^kx)$ and $g(x) = 1+x^p$.
Let $x_i = -1/\epsilon^i$, $i \in \{1, \ldots, p\}$.
We have $f(x_i) = 0$ for $i\in \{1, \ldots, p\}$. Other the other hand, $$g(x_i) = 1+(-1)^p\frac1{\epsilon^{ip}} = 1+(-1)^p.$$
For odd integer $p$, $g(x_i) = 0$. Since both $f$ and $g$ are $p$-th power polynomials, if they equal on $p+1$ ($f(0) = g(0)$) different points, then they must equal.