Show that the zeros of $\sum_{k=1}^n kz^{k-1}$ are inside the unit disc
Use the relation $$\frac{z^n-1}{z-1} = 1+z+z^2+\ldots+z^{n-1}$$ Your relation reduces to $$nz^n = 1+z+\ldots z^{n-1}$$ Let $|z| = r, r \in \Re$
Taking modulus, the above relation yields $$nr^n \leq 1+ r+r^2\ldots + r^{n-1}$$ $$nr^n-(1+r+r^2+\ldots r^{n-1}) \le 0$$
If $r>1$, $$f(r) = nr^n-(1+r+r^2+\ldots +r^{n-1}) = (r^n-1)+(r^n-r)+(r^n-r^2)+ \ldots+(r^n-r^{n-1})$$
Each of the bracketed terms $r^n-r^i,0\le i<n$ is positive. So, $f(r)>0$. There you have a contradiction. So, $r\le 1$.