Complex Finite Product $\prod_{k=0}^{n-1} (1-\zeta^k z)$ I am working on a review for a graduate level Complex Analysis course. The following problem is on the review:
Let $\zeta= e^{\frac{2\pi i}{n}}$ $(n\in \mathbb{N})$; show that 
$\displaystyle{\prod_{k=0}^{n-1} (1-\zeta^k z)=1-z^n}$
$\displaystyle{\prod_{k=1}^{n-1} (1-\zeta^k z)=1+z\ +...+\ z^n}$
I have proved it to myself for $n=3$ and recognized a "telescoping" behavior if you will. I know I have seen this formula before but can't remember where. Could someone please either point me in the right direction or help prove it? Thank you for your help.
 A: Just note this the first one nothing but the factorization of the of the polynomial 

$$z^n-1=0\,.  $$

Can you solve it? Here is how you find the roots

$$ z^n = e^{2\pi k i} \implies z = e^{2\pi k i/n}\, , \quad   k=0,1,2,\dots,n-1$$

You should be able to figure out the other one! 
A: Hints: For your first formula, notice that
$$(1-\zeta^kz) = (1-\overline{\zeta^{n-k}}z),$$
so that their product gives $1 - 2\Re(\zeta^k) + z^2$. Now you only have to separate the cases where $n$ is even or odd, and do some dirty work.
The second formula follows easily from the first by division of polynomials.
A: For the first formula, notice that $p(z)=1-z^n$ and $q(z)=\prod_{k=0}^{n-1} (1-\zeta^k z)$ are both polynomials of degree $n$ with exactly the same roots because $z^n=1$ iff $z^{-n}=1$.
To conclude that $p(z)=q(z)$ we need to prove that they have the same leading coefficient, $-1$. 
The leading coefficient $q_n$ of $q(z)$ is $(-1)^n \prod_{k=0}^{n-1} \zeta^k$.
But $\prod_{k=0}^{n-1} \zeta^k$ is the product of the roots of $p(z)$, and so is $(-1)^n\dfrac{1}{-1}=(-1)^n(-1)$ by Vieta's formulas.
Therefore, $q_n = (-1)^n (-1)^n (-1) = -1 = p_n$.
The second formula follows from the first:
$$
1-z^n
=\prod_{k=0}^{n-1} (1-\zeta^k z)
=(1-z)\prod_{k=1}^{n-1} (1-\zeta^k z)
$$
and so
$$
\prod_{k=1}^{n-1} (1-\zeta^k z)
=\frac{1-z^n}{1-z}=1+z\ +\cdots+\ z^n
$$
A: It was remarked that we can  easily derive the second formula from the
first  by  polynomial division,  therefore  we  will  prove the  first
formula here using what appears to be a somewhat exotic method, namely
the Polya Enumeration Theorem (PET).
 Given that $$\rho = \exp(2\pi i/n)
\quad\text{we seek to show that}\quad
f(z) = \prod_{k=0}^{n-1} (1-\rho^k z) = 1 - z^n.$$
Suppose  $Z(P_n)$   is  the  cycle   index  of  the   set  operator
$\mathfrak{P}_{=n}$ given by the recurrence by Lovasz which is
$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l+1} a_l Z(P_{n-l})
\quad\text{where}\quad
Z(P_0) = 1.$$
Now observe that
$$[z^q] f(z) = (-1)^q 
\left.
Z(P_q)(1+v+v^2+\cdots+v^{n-1})
\right|_{v=\rho}.$$
Next recall the OGF  of the set operator $\mathfrak{P}_{=q}$ which
is (this is the so-called exponential formula)
$$Z(P_q) = [w^q]
\exp\left(a_1 w - a_2 \frac{w^2}{2}
+ a_3 \frac{w^3}{3}
- a_4 \frac{w^4}{4}
+\cdots \right).$$
or
$$Z(P_q) = 
[w^q] \exp\left(\sum_{r\ge 1} (-1)^{r+1} a_r \frac{w^r}{r}\right).$$
Nóte that
$$\left.a_r\right|_{v=\rho}
= \left.(1+v^r+v^{2r}+\cdots+v^{(n-1)r})\right|_{v=\rho}
= n \times [[n|r]].$$
Therefore 
$$[z^q] f(z) = (-1)^q
[w^q] \exp\left(\sum_{r\ge 1} 
(-1)^{nr+1} n  \frac{w^{nr}}{nr}\right)
\\ = (-1)^q
[w^q] \exp\left(-\sum_{r\ge 1} 
\frac{(-w)^{nr}}{r}\right)
\\ = (-1)^q
[w^q] \exp\left(-\log\frac{1}{1-(-w)^n}\right)
= (-1)^q [w^q] (1-(-w)^n).$$
It now  follows that the constant  coefficient (on $q=0$)  is one, the
coefficients  on  $[z^q]$  where  $1\le  q\lt n$  are  zero,  and  the
coefficient  on $[z^n]$  i.e.  $q=n$  is 
$$(-1)^n  [w^n] (-(-w)^n)  = - (-1)^{2n} = -1$$ 
and hence
$$f(z) = 1 - z^n.$$
Commentary. The  recurrence for the  cycle index is  included here
for reference  and can  be used to  prove the exponential  formula and
vice versa. We have also made use  of the fact that for $r$ a positive
integer
$$\sum_{q=0}^{n-1} \rho^{rq} = n
\quad\text{when}\quad n|r.$$
On the other hand when $r$ is not a multiple of $n$ we get
$$\frac{\rho^{rn}-1}{\rho^r-1} = 0.$$
