The exercise is: Show that $\Phi_p(z) =1+z+z^2+z^3+...+z^{p-1}$
I started: If $p$ is prime, all the $p$th roots of unity are primitive (apart from 1) so can be expressed as $w, w^2, w^3,$ etc where $w=e^{i2\pi/p}$. In this form, the polynomial is $\Phi_p(z)=(z-w)(z-w^2)(z-w^3)...(z-w^{p-1})$. Using the fact that primitive roots come in cognate pairs, the product $w.w^2.w^3....w^{p-1}=1.$ For the $z$ term: $z(1+w+w^2+w^33+...+w^{p-1})$, I used the Hint (Ex 24) where I'd proved that $1+z+z^2+...+z^{n-1}=$$\frac{z^n-1}{z-1}$ to show that the sums of the powers of $w =-1$ as $z^n-1=0. $ For the higher powers of $z$, I argued from symmetry that all the powers of $w$ would be present in equal quantities, with a suitable number of $1$s for the odd powers and none for the even ones. But this is so weak. Is there a brilliant way of doing this that someone without even undergraduate level maths might understand (if they try really hard)? I looked at all the already answered questions but the ones I thought might answer this one I couldn't understand at all.
Gerry Myerson suggested I answer the question rather than delete it (to save it acquiring unmerited status as an 'unanswered question'). Based on the comment made by Claudius:
$z^p-1$is a polynomial, degree $p$, whose roots are exactly $1$ and the $p$-th roots of unity.
Factorised, $z^p-1=(z-1)(z-w)(z-w^2)(z-w^3)...(z-w^{p-1})$
${z^p-1}\over{z-1}$$=(z-w)(z-w^2)(z-w^3)...(z-w^{p-1})$ which is $\Phi_p(z)$
$\therefore \Phi_p(z)=1+z+z^2+z^3+...+z^{p-1} $ as required