Let $\lambda, \kappa$ be infinite cardinals. Does it follow from the inequality $\kappa\lt2^\lambda$ that $2^\kappa\lt2^{2^\lambda}$?
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The < sign seems not to work inside $LaTeX$, at least for me. \lt is recommended instead. – Ross Millikan Apr 20 '12 at 22:20

4Cross posted: http://mathoverflow.net/questions/94694/cardinalexponentation  if you cross post to MO please wait a day or two first. 30 minutes is simply unreasonable. – Asaf Karagila Apr 20 '12 at 23:07

You might want to have a look at [Easton's theorem](http://en.wikipedia.org/wiki/Easton%27s_theorem). It is mentioned also in this MO thread: [When 2^a = 2^b implies a=b (a,b cardinals)](http://mathoverflow.net/questions/17152/when2a2bimpliesababcardinals) and in the same question on this site [Does $2^X \cong 2^Y$ imply $X \cong Y$ without assuming the axiom of choice?](http://math.stackexchange.com/questions/74477/does2xcong2yimplyxcongywithoutassumingtheaxiomofchoice). – Martin Sleziak Apr 21 '12 at 06:45
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No.
Suppose $2^{\aleph_0}=\aleph_2$ and $2^{\aleph_1}=2^{\aleph_2}=\aleph_3$.
Let $\lambda=\aleph_0$ and $\kappa=\aleph_1$, now we have $2^\kappa=2^{2^\lambda}$.
Asaf Karagila
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