Answer
See explanations below.
Work Step by Step
Step 1. We have shown in the Exercise 97 that $1,w,w^2,w^3,...,w^{n-1}$ are the $n$ distinct roots of $1$ which means that $(w_k)^n=1$ where $k=0,1,2,...n-1$ is the index of $k$th solution.
Step 2. Given $z\ne0$ and $s$ is any $n$th root of $z$, we have $s^n=z$
Step 3. For any given term $sw_k$, $k=0,1,2,...n-1$, use the results from the above steps, we have $(sw_k)^n=s^nw_k^n=z\cdot 1=z$ which means that any term is a root of $z$ and the complete series of $n$ distinct terms form the $n$ distict roots of $z$
