Problem in solving a question related to roots of an equation. The question is :

Show that the equation $x^n+x^{n-1}+\cdots+x-1=0$ has unique positive root for all $n \in \mathbb {N}$ and all these positive roots lying in between $0$ and $1$ for all $n \geq 2$.

How can I solve it? Please help me. I just want to point out that I have solved the first part by Descartes' rule of sign. But I find difficulty in solving the remaining part of my question. So please give me a hint. Then I will retry it. Thank you in advance.
 A: So you have shown that a unique positive root exists for all $n$?  To get the bound on the roots, we can rearrange to give $x(x^{n-1} + x^{n-2} + \cdots + 1) = 1$.  It's impossible for $x \geq 1$ to solve this.
A: Note that $x=1$ is not a solution, so you can multiply by $(1-x)$ and get
$$
\left( {1 - x} \right)\left( {x^n  + x^{n - 1}  +  \cdots  + x} \right) - \left( {1 - x} \right) = \left( {1 - x} \right)x\left( {x^{n - 1}  + x^{n - 2}  +  \cdots  + 1} \right) - \left( {1 - x} \right) = x\left( {1 - x^n } \right) - \left( {1 - x} \right)\quad \left| {\;x \ne 1} \right.
$$
that is
$$
x\left( {1 - x^n } \right) = \left( {1 - x} \right)\quad  \Rightarrow \quad 0 = 1 - 2x + x^{n + 1} 
$$
A: Define $$f_n(x) = \frac{x^{n+1} - 1}{x-1}$$ with $f_n(1) = n+1$. This is a continuous function of $x$. 
Note that $x^n + x^{n-1} + \cdots - 1 = 0\ \ \text{iff} \ f_n(x) - 2 = 0  $
by the factorization of $x^{n+1} - 1$. 
For the latter equation, We have $f_n(0) = 1$ and $f_n(1) = n+1$. The result follows from the intermediate value theorem. 
To prove uniqueness, note, simply, that the given equation describes a function monotone increasing for positive $x$. 
