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?
 A: $f(0)=1$ and $f(1)=0$ for all $n$, so it suffices to show, for example, that $f(2/3)\lt0$.  But if $n\gt1$ then
$$f(2/3)=(2/3)^{n+1}-(4/3)+1\lt(2/3)^3-(1/3)=-1/27$$
A: if $x^{n+1} = 2x-1$, then $x^n = 2 - \frac{1}{x}$
Does this help?
A: Hint:
For $x \rightarrow 1^-$ write the function as $f(x)=x\left(x^n-2+\dfrac{1}{x}\right)$ and you can see that its value approximate $0$ by negative numbers.
A: Let's take $x = 1-\frac{1}{n+1}$ and prove that 
$$ x^{n+1}-2x+1<0 \Leftrightarrow \left(1-\frac{1}{n+1}\right)^{n+1}<2 \left(1-\frac{1}{n+1}\right)-1\Leftrightarrow \left(\frac{n}{n+1}\right)^n<1-\frac{1}{n}.$$
The last inequality can be proved if we notice that for $n=1$ it is true and the left side is decreasing function and the right side is increasing function.
A: 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.
A: Hint: If we recall that $a_{n+1}=\left(1-\frac{1}{n+1}\right)^{n+1}$ gives an increasing sequence converging towards $\frac{1}{e}$ and define $p_{n+1}(x)$ as $x^{n+1}-2x+1$, we have:
$$ p_{n+1}\left(1-\frac{1}{n+1}\right) < \left(\frac{1}{e}-1\right)+\frac{2}{n+1},$$
hence $p_{n+1}(x)$ has a root in $\left(0,1-\frac{1}{n+1}\right)$ for any $n\geq 2$.
A: Try to show that $f(a)<0$ for $a\approx1$ for example $a=\frac{9}{10}$
