Proof there is no largest cardinal number For any set $A$, I proved already that $|A| < |\mathcal{P}(A)|$ by injecting $A \to \mathcal{P}(A)$ by $a \mapsto \{a\}$ and then demonstrating, by contradiction, that no surjection can exist. I want to prove that there is no largest cardinal number, but the argument
$$|\mathbb{N}| < |\mathcal{P}(\mathbb{N})| < |\mathcal{P}(\mathcal{P}(\mathbb{N}))| < |\mathcal{P}(\mathcal{P}(\mathcal{P}(\mathbb{N})))| < \ldots$$
Perhaps more formally, I could suppose for the sake of contradiction that $|A|$ is the largest infinite cardinal, but this theorem implies that $|\mathcal{P}(A)| > |A|$, a contradiction.
How can I make this proof sufficiently formal?
 A: The simplest way to prove the claim - which you give - is to observe that $\vert A\vert<\vert\mathcal{P}(A)\vert$ for every $A$. (As an aside, this isn't really a proof by contradiction but rather a proof by negation, which is constructively valid.) On the other hand, the chain idea you write down doesn't work, first because it's not really clear what it means (what is "..."?), and second - and more importantly - because an unending chain of cardinals need not be unbounded.
That said, there are a couple comments worth making. First, there is an unbounded chain of cardinals coming from the powerset operation - namely, the $\beth$ numbers. Basically, we make precise the notion of "iterating the powerset $\alpha$-many times" for every ordinal $\alpha$; it's not hard to prove $\beth_\alpha\ge\vert\alpha\vert$ and so the chain of $\beth$ numbers is unbounded. This doesn't result in a new proof of the original claim, but it's an interesting thing to know about.
Second, and more subtly, the powerset might "miss things" - there could be cardinals in between $\vert A\vert$ and $\vert\mathcal{P}(A)\vert$. (To be precise, the usual axioms of set theory can neither prove nor disprove the existence of such gaps.) And on this note there is a genuinely different proof that there is no largest cardinal: namely, we can show that for every set $A$, there is a smallest cardinal $\kappa$ (or $\vert A\vert^+$) onto which $A$ does not surject, and moreover this $\kappa$ is in fact the cardinal successor of $\vert A\vert$ (that is, $\alpha>\vert A\vert$ and nothing is in between $\vert A\vert$ and $\alpha$). So this gives the second proof: replace "$\vert\mathcal{P}(A)\vert$" with "$\vert A\vert^+$." Of course this takes significantly more work to develop than the powerset version, but it is worth mentioning nonetheless.
A: Claim: There does not exist any set which has the largest cardinal. 
Proof: On the contrary let $A$ be the set with the largest cardinal. By Cantor's theorem, there exists no onto set from set $A$ to its power set $P(A)$ and therefore, $|A|\lt P(A)$, which is a contradiction. Proved.
