<s>[INST] Finding a singluar cardinality Given an infinite cardinal b such that $b^2=b$ I need to find a cardinal $x$ such that:
$$(x^b = 2^b)\quad  \land \quad (\forall c.(c>x)\to   (c^b>2^b))$$
Furthermore, I need to show there is only one such $x$.
I think the answer is  $ x = 2^b$ but have trouble proving the second part:$$(c>2^b) \to (c^b >2^b) $$ 
Any thoughts?
 [/INST] Certainly $x=2^b$ satisfies the given conditions: $x^b=\left(2^b\right)^b=2^{b^2}=2^b=x$, and if $c>x=2^b$, then $c^b\ge c>2^b$. To show uniqueness, suppose that $y$ is an infinite cardinal such that $y^b=2^b$, and $c^b>2^b$ whenever $c>y$. Then on the one hand we must have $y\le x$, since $y^b=2^b$, but on the other hand we must have $x\le y$, since $x^b=2^b$, so $x=y$.
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