Equation $x^{n+1}-2x+1=0$ 
Show without derivatives that the equation $x^{n+1}-2x+1=0,n>1,n\in\mathbb{R}$ has at least one solution $x_0\in (0, 1)$

I have thought that I need to use Bolzano for the equation as $f(x)$. So, we have: $$f(0)=1>0$$ $$f\left( \frac{1}{2}\right) =2^{-n-1}>0$$ $$f(1)=0$$
There is a problem here, as I cannot find any number or somehow use limit theorems in order for some $a\in (0, 1)$ to be $f(a)<0$. Any hint?

You want some $a \in(0,1)$ such that $f(a)<0$.
Writing $b =\frac{1}{a}$ this is equivalent to finding a $b >1$ such that 
$$
1-2b^n+b^{n+1} <0
$$
or 
$$2b^n > 1+ b^{n+1}$$
Now by Bernoulli we have
$$b^n >1+n(b-1)$$
So by choosing $n(b-1)>1$ you have $b^n>2$ and hence 
$$\frac{1}{2}b^n>1 \,.$$
Also, as long as $b <\frac{3}{2}$ you have
$$\frac{3}{2}b^n >b^{n+1}$$
Combining the two you get that for all $\frac{1}{n}+1< b < \frac{3}{2}$ we have
$$2b^n >1+b^{n+1}$$
This solution works for $n \geq 3$, as we need  $\frac{1}{n}+1< b < \frac{3}{2}$. For $n=2$ the inequalities are not strong enough, but in this case the problem is very easy to solve.
