Product identity: $x^{n+1}-1=\prod_{k=0}^{n}{(x-e^\frac{2ik\pi}{n+1}})$

Question

I recently stumbled upon this equation and frankly, I have no idea where this identity comes from. I tried to plot the two sides of the identity as a function and surprisingly, this equation holds. Wolfram Alpha hints that this has something to do with a “q-Pochhammer symbol”, but I have never heard of that before. $$x^{n+1}-1=\prod_{k=0}^{n}{(x-e^{\frac{2ik\pi}{n+1}})}{}$$ Does anyone know this identity and how do you prove it without difficult math? Thank you in advance.

Answer 1

HINT

By the foundamental theorem of algebra $$x^{n+1}-1=0$$

hase exactly $n+1$ roots on complex, given by

$$x=e^{\frac{2k\pi i}{n+1}}$$

for $k=0,...,n$.

Then recall that a polynomial can be factorized by its roots $r_i$

$$p_n(x)=(x-r_1)(x-r_2)...(x-r_n)$$

Answer 2

By fundamental theorem of algebra any monic polynomial $P_n(x)$ of degree $n$ can be represented as: $$ P_n(x)=\prod_{k=1}^n (x-x_k), $$ where $x_k$ are the roots of the polynomial.

In your case the roots of the polynomial $x^{n+1}-1$ are $x_k=e^{\frac{2\pi i}{n+1}k}$ with $k=0..n$.