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Consider the cardinals $\kappa:=2^{2^{\aleph_0}}$, $\lambda:=2^\kappa$, $\mu:=2^\lambda$. Which cardinal is larger, $\kappa^{\mu^\lambda}$ or $\lambda^{\mu^\kappa}$?

The only rules I believe I need are $2^c>c$ and $a.b=\max \{a,b\}$ for infinite cardinals $a,b,c$. I see that

$$\mu^\lambda = (2^{2^{2^{2^{\aleph_0}}}})^{2^{2^{2^{\aleph_0}}}} = 2^{(2^{2^{2^{\aleph_0}}}.2^{2^{2^{\aleph_0}}})} = 2^{2^{2^{2^{\aleph_0}}}}$$ and $$ \mu^\kappa = (2^{2^{2^{2^{\aleph_0}}}})^{2^{2^{\aleph_0}}} = 2^{(2^{2^{2^{\aleph_0}}}.2^{2^{\aleph_0}})} = 2^{2^{2^{2^{\aleph_0}}}} = \mu $$

so

$$ \kappa^{\mu^\lambda} = (2^{2^{\aleph_0}})^{2^{2^{2^{2^{\aleph_0}}}}} = 2^{2^{2^{2^{2^{\aleph_0}}}}} $$

and

$$ \lambda^{\mu^\kappa} = {(2^{2^{2^{\aleph_0}}})}^{2^{2^{2^{2^{\aleph_0}}}}} = 2^{2^{2^{2^{2^{\aleph_0}}}}}$$

but then I get that $\kappa^{\mu^\lambda} = \lambda^{\mu^\kappa}$, which means that either I or the question I'm doing is incorrect. Which one is it?

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    $\begingroup$ It's correct.${}$ $\endgroup$ – Simply Beautiful Art Apr 13 at 12:35
  • $\begingroup$ The question is not incorrect. The answer to “which is larger” simply is ”neither”. $\endgroup$ – celtschk Apr 13 at 14:03
  • $\begingroup$ @celtschk: There's more context to the question than I've given, which says that particular cardinals (of which I've provided two) can be ordered strictly in ascending order. $\endgroup$ – Adam Apr 13 at 15:21
  • $\begingroup$ Beware the trick question. Which is greater: One dozen or two six-packs? $\endgroup$ – DanielWainfleet Apr 19 at 19:52
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Your solution is correct, here is a more readable way of looking at it, remembering that $\mu>\lambda>\kappa$ by Cantor's theorem to simplify the products:

$$\mu^\lambda=(2^\lambda)^\lambda=2^{\lambda\cdot\lambda}=2^\lambda=\mu$$ $$\mu^\kappa=(2^\lambda)^\kappa=2^{\lambda\cdot\kappa}=2^\lambda=\mu$$

Now we're interested in comparing $\kappa^\mu$ and $\lambda^\mu$, but those are both $2^\mu$, since $\mu>\kappa$ and $\mu>\lambda$ (whenever you have two cardinals $\eta>\xi$, $\xi^\eta=2^\eta$).

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