Recursive seqence of power set, starting with the empty set

The operation that computes $|A_n|$ is known as tetration. It has many notations. Here I show its definition using two different notations: $$^na = \begin{cases} 1 &\text{if } n=0 \\ a^{\left(^{(n-1)}a\right)} &\text{if } n>0 \end{cases} $$ $$a\uparrow\uparrow n = \begin{cases} 1 &\text{if } n=0 \\ a^{a\uparrow\uparrow(n-1)} &\text{if } n>0 \end{cases} $$

Basically, tetration is a power tower where $a$ is repeated $n$ times.

According to Wikipedia, the first notation is used by Maurer[1901], Goodstein[1947] and Rudy Rucker[1982]. The second notation is known as Knuth's up-arrow notation[1976]. See this wiki about tetration: https://en.wikipedia.org/wiki/Tetration

Using tetration we have: $$|A_n| = {^{n-1}2} = 2\uparrow\uparrow(n-1)$$

It is interesting to note how fast this sequence grows: $$ \begin{align*} &|A_0| = {^{-1}2} = 0 \small \qquad \text{ (extending the definition)}\\ &|A_1| = {^02} = 1 \\ &|A_2| = {^12} = 2 \\ &|A_3| = {^22} = 4 \\ &|A_4| = {^32} = 16 \\ &|A_5| = {^42} = 65536 \\ &|A_6| = {^52} = 200352...9156736 \small \qquad \text{ (19729 digits in total in base 10)} \\ &|A_7| = {^62} = 2^{200352...9156736} \approx 10^{6.03\cdot 10^{19727} }\approx 10^{10^{19727.78}} \\ &... \end{align*} $$

As Michael Weiss noted in the comments:

Your answer is entirely correct. You've determined $|A_n|$; it's just an accident of mathematical history that there isn't a neater notation for the answer. The same is true for $A_n$.