Prove that for integral values of $n ≥ 1$, all the roots of equation $nz^n=1+z+z^2+\cdots+z^n$ lie within circle $|z|=\frac{n}{n-1}$.
My attempt:
Taking modulus on both sides, $n|z|^n=|1+z+...z^n|<=|1|+|z|+...|z|^n$
Dividing by |z|^n, we get, $n<=1+1/|z|+.....1/|z|^n$ $n<=1+1/|z|+\cdots$
Using infinite GP formula, if $|z|>1$ for this particular expression, $n<\frac{1}{1-1/|z|}$.
I could prove if $|z|>1$. How to prove for $|z|<1$
we could show that in the circle (boundary) $|z|=\frac n {n-1}$, we have $|1+z+\dots+z^n|\le 1+|z|+\dots+|z|^n=\frac{|z|^{n+1}-1}{|z|-1}\lt \frac{|z|^{n+1}}{|z|-1} =n|z|^n$ by Rouche's Theorem, we have that inside the circle $|z|\lt\frac n {n-1}$, polynomial $nz^n=1+z+\dots+z^n$ has same number of roots as polynomial $nz^n$. So all n roots of the polynomial is inside the circle