Prove that $\prod_{k=0}^{n-1}(z-\mathrm{e}^{2k\pi i/n})=z^n-1$ Prove that 
$$
\prod_{k=0}^{n-1}(z-\mathrm{e}^{2k\pi i/n})=z^n-1.
$$
In my some problem I have used 
$$
\prod_{k=0}^{7}(z-\mathrm{e}^{2k\pi i/8})=z^8-1.
$$
I have verified this.
So I think in general the following will be true
$$
\prod_{k=0}^{n-1}(z-\mathrm{e}^{2k\pi i/n})=z^n-1.
$$
Can someone help me to prove this general relation?
Thanks.
 A: Clearly, each of the $n$ DISTINCT complex numbers
$$
\lambda_k=\exp\left(\frac{2k\pi i}{n}\right), \quad k=1,\ldots,n,
$$
is a root of the polynomial
$$
p(z)=z^n-1.
$$
Hence, as we have found $n$ DISTINCT roots of $p$, which is a polynomial of degree $n$,
then we can analyze $p$ as
$$
p(z)=(z-\lambda_1)\cdots (z-\lambda_n)=\prod_{j=1}^n\left(z-\exp\left(\frac{2j\pi i}{n}\right)\right).
$$
A: I will elaborate the hint of @Liu Gang and the solution of @Yiorgos S. Smyrlis :
First, we need a standard result due to Bezout: Let $P(z)$ a polynomial. The remainder of the division of $P(z)$ by $(z-\lambda)$ equals $P(\lambda)$. Indeed, write
$$P(z) = (z-\lambda)Q(z) + R$$ where $R$ is the remainder, $\deg R < \deg (z-\lambda) = 1$ so $R$ is a constant. Substitute $z \mapsto \lambda$ in the equality and get 
$$P(\lambda) = 0 \cdot Q(\lambda) + R$$ and so $R = P(\lambda)$. 
As a consequence: $P(z)$ is divisible by $(z-\lambda)$ if and only if $P(\lambda) = 0$, that is, if and only if $\lambda$ is a root of $P(z)$. 
We'll show by induction on $n$ the following: let $P(z)$  and $\lambda_1$, $\ldots$, $\lambda_n$ distinct  so that $P(\lambda_1)= \ldots = P(\lambda_n) = 0$. Then we have the decomposition
$$P(z)= (z-\lambda_1) \ldots (z-\lambda_n) Q(z)$$ for some polynomial $Q(z)$. 
The case $n=1$ was treated above. 
Let's do the $n \to n+1$. 
Assume that it is true for $n$. Let $n\ge 1$ and $\lambda_1$, $\ldots$, $\lambda_{n+1}$ distinct so that $P(\lambda_1)= \ldots = P(\lambda_{n+1}) = 0$.  Then certainly  $P(\lambda_1)= \ldots = P(\lambda_{n}) = 0$. By the induction hypothesis we have 
$$P(z) = (z-\lambda_1) \ldots (z-\lambda_n) Q(z)$$ for some polynomial $Q(z)$. Plug in $z\mapsto \lambda_{n+1}$ in this equality. We get 
$$P(\lambda_{n+1}) = (\lambda_{n+1}-\lambda_1) \ldots (\lambda_{n+1}-\lambda_n) Q(\lambda_{n+1})$$
The LHS of this equality-- we know it's $0$. The RHS equals $0$ too but note that 
the factors $(\lambda_{n+1}-\lambda_1)$,  $\ldots$, $(\lambda_{n+1}-\lambda_n)$ are not $0$. It follows  that $Q(\lambda_{n+1})=0$ ( this is where we use the fact that we work in a field!) Now by the $n=1$ case ( Bezout) we get $Q(z) = (z-\lambda_{n+1}) S(z)$. Now we put all of it together to get the desired factorization for $n+1$. 
As a consequence: If $P(z) = a_n z^n + \ldots + a_0 $ is a polynomial of degree $\le n$ 
and $\lambda_1$, $\ldots$, $\lambda_n$ distinct  so that $P(\lambda_1)= \ldots = P(\lambda_n) = 0$ then 
$$P(z) = a_n (z-\lambda_1) \ldots ( z-\lambda_n)$$
Indeed, the quotient from the division of $P(z)$ by $(z-\lambda_1) \ldots ( z-\lambda_n)$ is of degree $\le 0$ so it must be $a_n$. 
We now note that the numbers $e^{i\ 2 k \pi/n}$, $0 \le k \le n-1$  are distinct and we have for $P(z) = z^n-1$ the equality $P(e^{i\ 2 k \pi/n}) = (e^{i\ 2 k \pi/n})^n -1 = e^ { i 2 k \pi}-1 = (e^{2 \pi i})^k-1 = 1-1=0$. How do we know they are distinct? Well we $\it{define}$ $2\pi$ to be the smallest number $>0$ with the property that $e^{i 2 \pi} = 1$.
