Find the least ordinal $β>0$ such that $αβ=β$ (given arbitrary $α>1$) The hint for this question was that $s=1+α+α^2+\cdots$ might be a candidate. While it's easy to check $αs=s$, it's not obvious how to check such $s$ is the least of all $β$ for which $αβ=β$.
Any hint would be appreciated :)
Update
Problem solved. Thanks go to Andrés E. Caicedo.
 A: Thanks for Andrés E. Caicedo's hint and Ross Millikan's suggestion:
Let $s=1+\alpha+\alpha^{2}+\cdots$. Then $s=1+\alpha s$. Given $\alpha>1$, $\alpha s=\alpha+\alpha^{2}+\cdots$ is no less than $\omega$ and thus $1+\alpha s=\alpha s$. 
It remains to show $s$ is the least ordinal which satisfies the equation $\alpha\beta=\beta$. We check that any $0<t<s$ is not a solution. Consider the partial sums of $s$: $s_{0} =1$, $s_{1} =1+\alpha$, ....
 Then $t<s$ implies there exists $n<\omega$ for which $t<s_{n}$. Suppose $t=\alpha t$, implying $t=\alpha^{k}t$ for all $k<\omega$. Thus $s_{n}>\alpha^{k}t\geq\alpha^{k}$ for all $k<\omega$, namely,
$s_{n}=1+\cdots+\alpha^{n}\geq\sup\left\{ \alpha^{k}|k<\omega\right\} =\alpha^{\omega}=\alpha^{(1+\cdots+n+1)+\omega}=\alpha^{1}\cdots\alpha^{n+1}\alpha^{\omega}$.  
This is impossible, for we can prove by induction that $1+\cdots+\alpha^{n}<\alpha^{1}\cdots\alpha^{n+1}$. Assuming for all $k<n$, $1+\cdots+\alpha^{k-1}<\alpha^{1}\cdots\alpha^{k}$, we have 
$1+\cdots+\alpha^{n-1}+\alpha^{n}<1+\cdots+\alpha^{n-1}+\alpha^{n+1}\leq\alpha^{1}\cdots\alpha^{n}+\alpha^{n+1}$. 
Let $1+\delta=\alpha^{n+1}$, then
$\alpha^{1}\cdots\alpha^{n+1}=\alpha^{1}\cdots\alpha^{n}(1+\delta)=\alpha^{1}\cdots\alpha^{n}+\alpha^{1}\cdots\alpha^{n}\delta>\alpha^{1}\cdots\alpha^{n}+\alpha^{1+\cdots+n}\geq\alpha^{1}\cdots\alpha^{n}+\alpha^{n+1}$
as expected.
