I do not understand the following passage/footnote in the book I am currently reading:
An initial ordinal $\lambda$ is called a strong limit cardinal if $2^\kappa < \lambda$ for every $\kappa < \lambda$. Note that this definition makes sense even in $ZF$.
Then, footnote:
If $\lambda$ is a strong limit cardinal, then $\mathcal P (\kappa) $ can be wellordered for every $\kappa < \lambda$. Thus, in $ZF$ alone one cannot prove the existence of an uncountable strong limit cardinal. But it is also relatively consistent with $ZF$ that an uncountable strong limit cardinal exists while the full Axiom of Choice fails.
What I understand: $\lambda$ is a cardinal, hence in particular an ordinal and hence well-ordered. If $2^\kappa < \lambda$ we therefore get a well-order on $\mathcal P (\kappa)$ induced by an order-isomorphism between $\mathcal P (\kappa)$ and some ordinal $\alpha < \lambda$.
What I don't understand: The sentence starting with "Thus...". How does it follow from $\mathcal P (\kappa)$ being well-ordered that in $ZF$ one cannot prove the existence of an uncountable strong limit cardinal? And: Would you show me how to prove the existence of an uncountable strong limit cardinal?
Many thanks for your help.