How to prove the following function has only one root in interval [0,1]:
$f(x)=(1-x^b)^b-(1-x)$ where b>1 is a positive integer
Some hint so far: we can easily prove that $g(x)=1-x^b$ has exactly one fixed point in [0,1] If we can proof that $g(g(x))=1-(1-x^b)^b$ has exactly one fixed point in [0,1] then the result follows.
Hint for easier solution: $f(0) = f(1) = 0$ and $f''$ has only one root in $(0,1)$.