How to solve this? (question involving nth roots of unity)?

Question

If 1, w, w^2, w^3,.........,w^(n-1) are the n, nth roots of unity, then (1-w)(1-w^2)(1-w^3)...(1-w^(n-1)) equals?

Answer 1

(x−1)(x−ω)(x−ω²) … (x−ωⁿ⁻¹) = xⁿ−1 = (x−1)(xⁿ⁻¹+xⁿ⁻²+ … +x+1)



∴ (x−ω)(x−ω²) … (x−ωⁿ⁻¹) = xⁿ⁻¹+xⁿ⁻²+ … +x+1 for all x



Then just set x=1 to get (1−ω)(1−ω²) … (1−ωⁿ⁻¹) = n

Answer 2

It looks like the answer is n

Consider a few examples, distinguishing roots for various n with a, b c etc

n = 3:

a = e^(i2π /3) = [-1 + √(3)i]/2

a^2 = e^(i4π /3) = [-1 - √(3)i]/2

a^3 = e^(i6π /3) = 1

(1 - a)(1 - a^2) = 1 – a – a^2 + a^3

Use – (a – a^2) = 1 = a^3

(1 - a)(1 - a^2) = 3



n = 4: b = i, b^2 = -1, b^3 = -1, b^4 = 1

(1 - b)(1 - b^2) (1 – b^3) = 1 – b – b^2 + b^4 + b^5 – b^6

(1 - b)(1 - b^2) (1 – b^3) = 1 – i + 1 + 1 + i + 1 = 4



n = 5:

c = e^(i2π /5)

c^2 = e^(i4π /5)

c^3 = e^(i6π /5)

c^4 = e^(i8π /5)

(1 - c)(1 - c^2) (1 – c^3)(1 – c^4)= 1 – c – c^2 + 2c^5 – c^8 – c^9 + c^10

Note that c^5 = c^10 = 1, so that becomes 1 – c – c^2 + 2 – c^8 – c^9 + 1

c^8 = c^3, and c^9 = c^4, remaining terms are – c – c^2 – c^3 – c^4 = 1

(1 - c)(1 - c^2) (1 – c^3)(1 – c^4) = 5



But that is an heuristic verification; not really a proof.

Perhaps J or someone else will add to this by generalising why the product is always n

Answer 3

product_(m = 1)^(n - 1) (1 - (w^m)) = 1/gamma(2 - (w^n))

Answer 4

n