# There is a stright formula for calculate n Power Sets P(A) of a A set with m members

By using Cantor's Theorem: if We have a Set A, with n members the Cardinality of the Power Set is $P(A)=2^n$. There is Formula to calculate n Power Sets $P(...P(P(A))...)$ of a set A with m elements without doing the simple iteration $(((2^2)^2)..)^2^m$ n times? Many thanks

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What you wrote is incorrect: the cardinality of the $n$th iterate of the power set operation on a set of size $m$ is $2^{2^{2^{\dots{^{2^m}}}}}$ where we have a tower of $n$ $2$s. –  Arthur Fischer Aug 16 '12 at 8:15
You are right I forgot to put the m in the end –  Hernan Aug 16 '12 at 8:25
It's not just the missing $m$, it's the order of operations. This is essentially part of Brian's answer below, but if $|A| = m$, then $| P(A) | = 2^{|A|} = 2^m$, and $|P(P(A))|=2^{|P(A)|}=2^{2^m}$; etc. If $m = 3$, then $|P(P(A))| = 2^{2^3} = 2^8 = 256$, while your formula gives $(2^2)^3 = 4^3 = 64$. –  Arthur Fischer Aug 16 '12 at 8:29

The correct formula is $$\large{2^{\left(2^{\left(2^{\dots(2^n)}\right)}\right)}}\;.$$ That is, you start with $2^n$ for the cardinality of $\wp(A)$ when $|A|=n$, then get $2^{2^n}=2^{(2^n)}$ for the cardinality of $\wp(wp(A))$, and so on. I know of no simpler form for this iterated power.

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The term $(((2^2)^2)\ldots)^2$ simplifies to $2^{2^n}$.This follows from the general identity $(x^m)^n=x^{m\cdot n}$ and induction.

However, what you should actually be computing (as S4M points out in the comments) is an expression like $$2^{2^{2^{2^{2^n}}}}$$ which does not simplify any further.

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