Defining strong limit cardinals in $ZF$ 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.
 A: The existence of a strong limit would prove the existence of a well-order of the reals, because it has to be greater than $2^{\aleph_0}$ and the first passage would apply.  Since we know $ZF$ doesn't prove such a well-order exists, there cannot be a proof that a strong limit exists.
A: It is consistent that the axiom of choice fails and $2^\omega$ cannot be well-ordered. It follows that no power set of an ordinal can be well-ordered, and so no limit cardinals are strong limit cardinals.
An interesting paper on the topic is:

Blass A., Dimitriou I. M. , and Löwe B. Inaccessible cardinals without the axiom of choice. Fundamenta Mathematicae , vol.194, pp. 179-189 

Which can be found here.

In ZFC to show that there is an uncountable strong limit cardinal, remember the definition of $\beth$ numbers and consider $\beth_\delta$ for any limit ordinal $\delta$.
As for the last remark about consistency, we can force that the axiom of choice fails above the first strong limit cardinal, $\kappa=\beth_\omega$. In this case all the cardinals up to $\beth_\omega$ and their power sets can be well-ordered and so $\kappa$ is still a strong limit cardinal but the axiom of choice fails in general.
