Is there a highest order of infinity? Does there exist an infinite set of cardinality such that it can never be reached by taking power sets of a set with cardinality aleph-null. Please prove your answer, or include a link to a proof. I apologize for any excessively loose terminology, I am new to this subject of different infinities.
 A: The answer to the question in the title is no: There is no highest cardinality: Given any set, its power set is always larger. In fact, given any collection $I$ of sets, if we take their union and then the power set of the result, we get a set of cardinality larger than that of any set in $I$. 
The answer to the question in the body is yes: We need a definition. The hierarchy of beth ($\beth$) numbers is defined by transfinite recursion: First, $\beth_0=\aleph_0$ is the size of the natural numbers. Given $\beth_\alpha$, we define $\beth_{\alpha+1}$ as the size of its power set. Finally, given a limit ordinal $\lambda$, we define $\beth_\lambda$ as the supremum of the $\beth_\beta$ for $\beta<\alpha$, that is, as the size of the union of the set $\{\beth_\beta\mid \beta<\alpha\}$. (Here I am identifying cardinals with sets of ordinals, so for any cardinal $\kappa$, the set $\kappa$ itself has indeed cardinality $\kappa$.) For example, $\omega=\omega_0$ is the smallest (infinite) limit ordinal, and $\beth_\omega=\sup\{\beth_n\mid n\in\mathbb N\}=\bigcup_n\beth_n$.
In terms of this hierarchy, a set has size that cannot be reached by taking repeated power sets of $\mathbb N$ iff its cardinality is $\beth_\omega$ or larger: A set has size not reachable by taking repeated power sets of $\mathbb N$ iff its size is larger than $\beth_0=|\mathbb N|$, $\beth_1=|\mathcal P(\mathbb N)|$, $\beth_2=|\mathcal P(\mathcal P(\mathbb N))|$, etc, that is, iff its size is at least the supremum of the $\beth_n$ (which is precisely what $\beth_\omega$ is). Of course, any larger cardinality (such as $\beth_\omega^+$, $\beth_{\omega+1}$, etc.) is not reached either.
The proof that sets of this size exist requires a modicum of attention: By the axiom of replacement, we can show that the set $\{\beth_n\mid n\in\mathbb N\}$ exists, since this is the image of $\mathbb N$ under the definable map $n\mapsto|\mathcal P^n(\mathbb N)|$. The union axiom ensures now that $\bigcup_n \beth_n$ exists, and this is exactly $\beth_\omega$. The use of replacement is essential, since $V_{\omega+\omega}$ is a model of the theory $\mathsf{ZC}$, Zermelo-Fraenkel without replacement, and in this model any set has size bounded by some iterated power set $\mathcal P^n(\mathbb N)$.
Just about any standard text in set theory should have a proof of this result. I suggest Moschovakis's Notes on set theory. The book also discusses the cumulative hierarchy of the $V_\alpha$, to which the set $V_{\omega+\omega}$ of the previous paragraph belongs. This hierarchy is defined by setting $V_0=\emptyset$, $V_{\alpha+1}=\mathcal P(V_\alpha)$ for all ordinals $\alpha$, and $V_\lambda=\bigcup_{\beta<\lambda}V_\beta$ for all limit ordinals $\lambda$.
Of course, we could change the question and allow instead transfinite iterations of the power set function, so we look not just at $\mathbb N,\mathcal P(\mathbb N),\dots,\mathcal P^n(\mathbb N),\dots$, but we continue the sequence by looking at $\mathcal P^\omega(\mathbb N)=\bigcup_n\mathcal P^n(\mathbb N), \mathcal P^{\omega+1}(\mathbb N)=\mathcal P(\mathcal P^\omega(\mathbb N)),\dots,\mathcal P^\alpha(\mathbb N),\dots$ where the sequence extends over all ordinals $\alpha$. Now the answer to the question is again no: Note that $|\mathcal P^\alpha(\mathbb N)|=\beth_\alpha$. Any set has size bounded by some $\beth_\alpha$. Again, any standard text should explain why this is the case.
