I am self studying number theory from Tom M Apostol Introduction to analytic number theory and I got struck on this problem in Ch-14 (problem 11 (a)).
Let $\epsilon =e^{2\pi i/k}$ where k$\geq$1 . Show that for all x we have $\prod_{h=1}^{k} (1-x{\epsilon}^h)=1-x^k$ .
I wrote that identity to be proved is equal to $x^k-1={(-1)^k}\prod_{h=1}^{k} ({x\epsilon}^{h}-1)$.
Now I wrote $ x^k -1 = \prod_{h=1}^k (x-{\epsilon}^h) = {\epsilon}^{-k (k+1)/2}\prod_{h=1}^k (x{\epsilon}^{-h}-1)$.
I am not able to move foreward. Can you please help .
Thanks!!
For finding the product of some factors, you can use Vieta's formulas, if you can find a polynomial whose roots are those numbers precisely.
In our case, we can use polynomial transformations, along with the fact that $\epsilon^h , h=1,...,k$ are exactly the roots of $y^k - 1 = 0$.
From this fact, $x\epsilon^h , h=1,...,k$ are exactly the roots of $y^k - x^k = 0$.
From there, the quantities $(x\epsilon^h-1) , h=1,...,k$ are exactly the roots of $(y+1)^k - x^k = 0$. Therefore, the product of all roots i.e. $\prod_{h=1}^k(1-x\epsilon^h)$ is the constant term of this polynomial with a $(-1)^k$ factor, which from the binomial expansion is just $(1-x^k)(-1)^k$. Finally, the $(-1)^k$ from both sides cancel by reversal of signs in the product (you have already shown this in your working), to give the result.