We know $P(x) = ax^n + bx^{n-1} + cx^{n-2} + ... + z = 0$ .
also $ \alpha , \beta , \gamma$ and ... $\omega $ are roots of $P(x)$.
in the other hand :
$S_n = \alpha^n + \beta^n + \gamma^n + ...+ \omega^n $ , $S_{n-1} = \alpha^{n-1} + \beta^{n-1} + \gamma^{n-1} + ...+ \omega^{n-1} $ and so on.
How we can prove this relation by induction ?
$aS_n + bS_{n-1} + cS_{n-2} + .... + zS_0 = 0$
It is not clear how to use the induction and in fact we don’t need it, becuse we can proof the claim directly. For the simplicity we denote the polynomial $$P(x)=a_0+a_1x+\dots a_n x^n$$ and its roots as $$\alpha_1,\dots,\alpha_n.$$ Then for each $i$ $$S_i=\sum_{j=1}^n \alpha_j^i.$$ If we additionally put $S_0=n$ we obtain
$$\sum_{i=0}^n a_iS_i=\sum_{i=0}^n a_i\sum_{j=1}^n \alpha_j^i=\sum_{j=1}^n\sum_{i=0}^n a_i \alpha_j^i =\sum_{j=1}^n P(\alpha_j)= \sum_{j=1}^n 0=0.$$
PS. We have to have $nz$ instead of $z$ in the initial formulation of your question.