Prove the uniqueness of steady state I have a difference equation 
  $$ 
p_t^{1-\alpha}=\alpha\sigma \left(y-p_t-\frac{(\sigma p_{t-1}^\alpha+b)p_t^{1-\alpha}}{\alpha\sigma} \right) 
$$
  where $\alpha \in [0,1]$ and everything else is $&gt;0$. 


I need to prove that this equation has a 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?
Rearranging the steady state equation
$$
\overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}-
\alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma}
$$
we get
$$
(1 + \alpha)\overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}-
\frac{a+1}{\sigma}.
$$
As $\alpha \in [0,1]$, the left hand side of the equation is increasing in $\overline{p}$ and the right hand side is decreasing. At least one of these is strictly monotonic because $\alpha$ cannot be $0$ and $1$ at the same time. Hence at most one solution is possible.