I'm sorry in advance for how specific this problem is. I've been trying to make it as generic as possible but this is as far as I could get. I want to prove that a fixed point is attractive.
I didn't find much on Google about proving the attractiveness of a fixed point, but if there is a general methodology, that would be very helpful already.
The function is: $$ f(x) = (ax + b)^{1/\alpha} $$ with $a>0$, $\alpha\in(0,1)$ and $b\in(0,b_0)$ where $\displaystyle{b_0 = (1-\alpha)\left(\frac{\alpha}{a}\right)^{\alpha/(1-\alpha)}}$. The domain of this function is $[0,x_0)$, with $\displaystyle{x_0 = \left(\frac{\alpha}{a}\right)^{1/(1-\alpha)}}$.
I proved that this function has one unique fixed point in its domain, but I don't know how to prove that it is attractive. Specifically, the derivative is not absolutely lesser than 1. What I know already about the derivative is that it is strictly positive and increasing, and takes values in $[0,x_0)$.
EDIT:
I made a mistake in my calculations; the derivative takes values in $[0,1)$ which is sufficient to prove the fixed-point is attractive. Thanks to Julian for his answer.
We will work on $[0,+\infty[$.
First observe that for $x\geq 0$: $$ f(x)=x\quad\Leftrightarrow\quad ax+b=x^\alpha. $$ Now draw the graphs of $ax+b$ and $x^\alpha$ to convince yourself that there could be $0$, $1$ or $2$ positive fixed points in general.
It turns out that the condition $b<b_0$ guarantees that there are $2$ positive fixed points.
Indeed, consider the point $$ x_2:=\frac{\alpha b}{(1-\alpha)a}. $$ At this point, we have, after simplifications: $$ ax_2+b<x_2^\alpha\quad\Leftrightarrow\quad b< (1-\alpha)\left( \frac{a}{\alpha}\right)^\alpha=b_0. $$ The latter is one hypothesis given by the OP, so now we know why it is here. Since the graph of $x^\alpha$ is above $ax+b$ at $x_2$ whereas it is below at $0$ and at $+\infty$, we know that there are two fixed points $x_1,x_3$ such that $0<x_1<x_2<x_3$.
Next compute the derivative at $x_1$ and use the fixed point condition: $$ f'(x_1)=\frac{a}{\alpha}(ax_1+b)^{1/\alpha-1}=\frac{ax_1}{\alpha(ax_1+b)}. $$ Finally, we see after simplification that: $$ f'(x_1)<1\quad\Leftrightarrow\quad x_1<x_2. $$ So the smallest fixed point is attractive.
Note that the last paragraph is actually the consequence of the more general observation made by Will Jagy.
Finally, the condition $x_1<x_0$ is also fulfilled. Indeed, we have $ax_0+b<x_0^\alpha$ so $x_1<x_0$, since the former turns out to be equivalent to $b<b_0$ again.