Let $X$ be an infinite set, $F$ the set of filters and $U$ the set of free ultrafilters on $X$. Then for the cardinalities we have $|U| \leq |F| \leq 2^{2^{|X|}}$. I have found several proofs that we have equality throughout. But I don't understand them. So please give an explaination of this fact.
The powerset of $X$ consists of pairs of set and its complement. A free ultrafilter contains either the set or its complement. So the number of combinations is the binomial coefficient of $2^{|X|} $ over $2^{|X| - 1}$ which equals $\frac{2^{|X|}!}{2^{|X| - 1}! \times 2^{|X| - 1}!}$. How do I use cardinal arithmetic to show that this equals $2^{2^{|X|}}$? Or is this line of thought not leading anywhere?