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?

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 $n\geq 2$.
