Proving the identity $ \prod_{k=1}^{n-1}(1 - e^{\frac{2\pi ik}{n}}) = n$

Question

I'm trying to prove this identity involving roots of unity, and would like to know if the following is a valid chain of reasoning.

Looking at the complex polynomial solved by the n-th roots of unity we have:

$$ w^n - 1 = (w - w_0)(w - w_1)...(w - w_{n-1}) = (w - 1)(w - w_1)...(w - w_{n-1}) $$

where $w_0, w_1, ..., w_n$ are the n-th roots of unity (i.e $w_k = e^{i\frac{2\pi k}{n}}$, with $w_0 = 1$).

At the same time, we have the telescoping sum identity:

$$ w^n -1 = (w - 1)(w^{n-1} + w^{n-2} + .... + w + 1) $$

therefore, we can cancel out the factor $(w-1)$ and get the identity:

$$ (w - w_1)...(w - w_{n-1}) = (w^{n-1} + w^{n-2} + .... + w + 1) $$

Whereupon setting $w = 1$ we get the identity:

$$ (1 - w_1)...(1 - w_{n-1}) = (1 - e^{i\frac{2\pi}{n}})...(1 - e^{i\frac{2\pi (n-1)}{n}}) = (1^{n-1} + 1^{n-2} + .... + 1^1 + 1) = n $$

Is this kosher? specifically I'm a little uncomfortable about the dividing out of the factor $w - 1$ and then setting $w = 1$, since that would've made the factor $w - 1$ zero!

Answer 1

You've done it very well! We have $$(w - 1)(w^{n-1} + w^{n-2} + \cdots + w + 1)=w^n-1=(w - 1)(w - w_1)\cdots(w - w_{n-1})\tag{$*$}$$ for all $w\in\mathbb C$. We denote $f(w)=w^{n-1} + w^{n-2} + \cdots + w + 1, g(w)=(w - w_1)\cdots(w - w_{n-1})$ for $w\in\mathbb C$, then clearly $f$ and $g$ are continuous in $\mathbb C$.

Now, $(*)$ implies that $(w-1)f(w)=(w-1)g(w)$ for all $w\in\mathbb C$, hence $$f(w)=g(w)\qquad \forall w\in\mathbb C\setminus\{1\}.\tag{$**$}$$ It follows from $(**)$ and the continuity of $f,g$ that $f(1)=g(1)$, hence $$n=f(1)=g(1)=\prod_{k=1}^{n-1}(1 - e^{\frac{2\pi ik}{n}}).$$

Answer 2

Yes, this is kosher and is "the" way to do it.

Think of these as formal polynomials in $\Bbb C[w]$. Let $p(w)=1+w+\cdots+w^{n-1},q(w)=w-1,h(w)=(w-w_1)\cdots(w-w_{n-1})$. You know that $pq=hq$ and that $\Bbb C[w]$ is an integral domain, $q$ nonzero; $p=h$ follows. More concretely, this would boil down to a "comparing coefficients" argument. Essentially, polynomial division works for the same reason integer division works, so long as you divide by a nonzero element. Thinking in $\Bbb C[w]$ land, $(w-w_1)\cdots(w-w_{n-1})=\frac{w^n-1}{w-1}=1+w+\cdots+w^{n-1}$ is an equation of polynomials; there is no worry about "what if $w=1$?" since $w$ is not actually a complex number.

If you want some analysis over some algebra, just note that you can (now viewing these polynomials as functions $\Bbb C\to\Bbb C$) say $p(w)=h(w)$ for all $w\neq1$ and since polynomials are continuous it follows $p(1)=h(1)$ as well.

Answer 3

In effect, the division by the term $w-1$ implicitly assumes that $w$ is different from $1$ from now on. You can solve this problem by going to the limit:

$$n=\lim_{w\rightarrow 1}\frac{w^{n}-1}{w-1}= \lim_{w\rightarrow 1} \prod_{k=1}^{n-1}(w-w_{k})=\prod_{k=1}^{n-1}\left(1-e^{ \frac{2\pi i k}{n} } \right)$$