Proving $k^2 \leq 2^{2^{k}}$ from Kuratowski's definition of ordered pair. Is it possible to extend this? If we use Kuratowski's definition of an ordered pair, we have that $\left(a,b\right)=\left\{\left\{a\right\},\left\{a,b\right\}\right\}$, where $a,b\in X$ and $\left(a,b\right)\in X\times X$. However, note that this definition also implies that $\left(a,b\right)\in\mathscr{P}\left(\mathscr{P}\left(X\right)\right)$.
Suppose that $X$ is a finite set with $k$ elements. Then, it's not hard to prove that $|X\times X|=k^{2}$ and $|\mathscr{P}\left(X\right)|=2^{k}$, which gives us $|\mathscr{P}\left(\mathscr{P}\left(X\right)\right)|=2^{2^{k}}$. Given that this definition of ordered pairs implies that $X\times X\subseteq \mathscr{P}\left(\mathscr{P}\left(X\right)\right)$, that would make this a valid proof that $k^{2}\leq 2^{2^{k}}$ for all integers $k\geq1$.
Question: Can this definition be extended to ordered $n$-tuples to prove similar inequalities for $k^{n}$? If so, what does it looks like? If not, why not?
I know that we can recursively define ordered triples by
$$\left(a,b,c\right)=\left(\left(a,b\right),c\right)=\left\{\left\{\left(a,b\right)\right\},\left\{\left(a,b\right),c\right\}\right\}$$
$$=\left\{\left\{\left\{\left\{a\right\},\left\{a,b\right\}\right\}\right\},\left\{\left\{\left\{a\right\},\left\{a,b\right\}\right\},c\right\}\right\},$$
and so on for ordered $n$-tuples. However, this definition means that $\left(a,b,c\right)$ is not the member of any level of power set of $X$ (not $\mathscr{P}\left(X\right)$, $\mathscr{P}\left(\mathscr{P}\left(X\right)\right)$, $\mathscr{P}\left(\mathscr{P}\left(\mathscr{P}\left(X\right)\right)\right)$, etc.), which means that we can't use (a slightly modified form of) the original argument to find an inequality for $k^{3}$.
Since $2^{5} > 2^{2^{2}}$, this means that the ordered $5$-tuples of elements of $X$ can't be contained inside of $\mathscr{P}\left(\mathscr{P}\left(X\right)\right)$ (at least for $X$ with at least $2$ elements), which means defining higher-order ordered tuples that are consistent with the original Kuratowski definition but also extend it in a way that allows for further power set-related inequality arguments will require a deeper nesting of power sets, if such a definition exists.
Note: The inequalities that would be found using this would be terribly loose to the point of being beyond trivial. Since there are no integers $k\geq1$ with $k^{3}$ or $k^{4}$ greater than $2^{2^{k}}$, the further nesting of exponential functions that will result from the nesting of power sets needed to define ordered triples and ordered $4$-tuples will already lead to inequalities that are extremely weak.
 A: Well, you could define $(a,b,c)$ as $((a,b),(c,c))$, and then it is an element of $\mathscr{P}^4(X)$.  More generally, if $n\leq 2^m$, you can represent $n$-tuples as a tree of $m$-tuply nested ordered pairs, and thus as elements of $\mathscr{P}^{2m}(X)$.  This shows that $$k^n\leq f^{2\lceil \log_2 n\rceil}(k)$$ where $f(x)=2^x$.  (Of course, this bound can also be obtained by just iterating $k^2\leq 2^{2^k}$ to get $k^4=(k^2)^2\leq (2^{2^k})^2\leq 2^{2^{2^{2^k}}}$ and so on.)
You can do better at least in some cases, though.  For instance, you can represent triples in $\mathscr{P}^3(X)$ by defining $(a,b,c)$ as $\{(a,b),(a,c),(b,c)\}$.  More generally, you can represent an $n$-tuple as the set of all the $(n-1)$-tuples obtained by deleting one entry in the $n$-tuple (for any $n>2$), so inductively this represents $n$-tuples as elements of $\mathscr{P}^n(X)$.  This gives the bound $$k^n\leq f^n(k),$$ which is better than the previous bound for $n=3$ and $n=5$.  Or, you can use the earlier representation of $2^m$-tuples as elements of $\mathscr{P}^{2m}(X)$ to represent $(2^m+1)$-tuples as elements of $\mathscr{P}^{2m+1}(X)$, which is slightly better than the $\mathscr{P}^{2m+2}(X)$ that would be given by the first method.
To prove that this works, let $s$ be an $n$-tuple and let $R$ be the set of $(n-1)$-tuples obtained by removing an entry from $s$; we will recover $s$ from $R$.  If all elements of $R$ have the same first entry (say, $a$), then $a$ must be the first entry of $s$.  Moreover, there is then a unique element of $R$ which starts with fewer $a$s than every other element of $R$ (namely the $(n-1)$-tuple obtained by removing the first entry of $s$), and that element is the remaining $n-1$ entries of $s$.
So, we may assume that not all elements of $R$ have the same first entry.  If there is some $a$ such that two different elements of $R$ start with $a$, then $a$ must be the first entry of $s$ and the remaining entries are given by the unique element of $R$ that doesn't start with $a$.  Thus, we may assume that $R$ has only two different elements, say one starting with $a$ and another starting with $b$.  But this means the entries of $s$ can only be $a$ and $b$ (if $s$ had three distinct entries, they would give three distinct elements of $R$).  Moreover, all the $a$s must be consecutive, since removing $a$s in different consecutive blocks would give different elements of $R$, and similarly the $b$s must be consecutive.  We can count how many $a$s there are in $s$ (the maximum that there are in any element of $R$) and similarly for the $b$s, and we can tell whether the $a$s come first or the $b$s come first since $n>2$.  Thus, we can recover $s$ from the set $R$.
For $n=3$, at least, this is optimal in the following sense: it is impossible to represent ordered triples as elements of $\mathscr{P}(\mathscr{P}(X))$ (by a formula that sends a triple $(a,b,c)$ to some doubly nested set expression in $a,b,$ and $c$).  Clearly such a representation of a triple $(a,b,c)$ would have to involve all three of $a,b,$ and $c$.  But now consider the 6 triples $(a,a,b),(b,b,a),(a,b,a),(b,a,b),(b,a,a),(a,b,b)$.  Each of these must be represented by a distinct element of $\mathscr{P}(\mathscr{P}(\{a,b\}))$ which is not fixed if you swap $a$ and $b$.  Thus, each one must contain exactly one of $\{a\}$ and $\{b\}$.  This means that each one also must contain $\{a,b\}$, since the formula for $(a,b,c)$ must involve all three of $a,b,$ and $c$.  But now we have a problem: there are only 4 different subsets of $\mathscr{P}(\{a,b\})$ satisfying this constraint, so our 6 triples cannot all be distinct.
