Question on primitive roots of unity

Solution 1:

Recall the formula for the summation $1+x+x^2+\cdots+x^n = {x^{n+1}-1 \over x-1}$ when $x\neq 1$.

In particular, this shows that $f(x)=x^{p}-1 =(1+x+x^2+\cdots+x^{p-1})(x-1)$.

Since $\omega$ is a root of unity, we have $f(\omega^k) = 0$ for $k=0,...,p-1$.

Since $\omega$ is primitive, $\omega^0,...,\omega^{p-1}$ are distinct and so $f(x) = (x-1)\cdots (x-\omega^{p-1}) $.

Hence $1+x+x^2+\cdots+x^{p-1} = (x-\omega^1)\cdots (x-\omega^{p-1}) $.

Solution 2:

Hint: what does $x^n-1$ look like factored?

Solution 3:

Let $f(x)=x^p-1$ and $g(x)=f(1-x)$. Then $g(1-\omega^k)=f(\omega^k)=0$ for all $k$.

$(1-\omega)(1-\omega^2)\cdots (1-\omega^{p-1})$ is thus the product of all roots of $g$ except $0$. So, expand $g$, factor $x$, and look at the independent term.