Proving uniqueness of a steady state I have a difference equation 
$$
p_t^{1-\alpha}=\alpha\sigma(y-p_t-\frac{(\sigma p_{t-1}^\alpha+b)p_t^{1-\alpha}}{\alpha\sigma})
$$
where $\alpha \in [0,1]$ and everything else is >0. 
I need to prove that this equation has unique steady state.
This is what I have done so far; 
Simplified the expression to write it in the closed form as follows;
$$
p_{t-1}=\left[\frac{\alpha y}{p_{t}^{1-\alpha}}-\alpha p_{t}^{\alpha}-\frac{a+1}{\sigma}\right]^{1/\alpha}
$$
Substituted $p_{t-1}=p_t=\overline{p}$ in the closed form, this gave. 
$$
\overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}-
\alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma}
$$
I'm stuck here. How can I prove that $\overline{p}$ has a unique solution? 

Well, what you want to study is the existence of solutions for
$$
f(p;\alpha) = (1+\alpha)p^{\alpha} - \alpha y p^{\alpha-1} = - \frac{a+1}{\sigma}.
$$
Where, as you've said, $\alpha \in [0,1]$, $y,\,a,\,\sigma >0$. First, is clear that for $\alpha = 0$, there is no solution for $p>0$ (why?). If $\alpha = 1$, then there is a solution if and only if $\alpha y - \frac{a+1}{\sigma} > 0$$.
Now, let $\alpha \in (0,1)$. 
$$
\text{What can you say about } \lim_{p\to -\infty} f(p;\alpha) \,\text{ and } \lim_{p\to \infty} f(p;\alpha)\text{?}
$$
$$
\text{What about }f'(p;\alpha)\text{?}
$$
