Cardinal exponentiation My understanding of exponentiation of cardinals leads to the conclusion that if $2 \leq \kappa \leq \lambda$, then $2^\lambda = \kappa^\lambda$
, because:
$2^\lambda \leq \kappa^\lambda \leq (2^\kappa)^\lambda =  2^{\kappa \cdot \lambda} = 2^\lambda $
However, I am slightly confused at this because Cantor’s Theorem states that $\lvert A \rvert < \lvert \mathcal{P}(A) \rvert $, for every set A.
Hence, $\kappa < 2^\kappa $, for every cardinal $\kappa$
I know it would lead to the inconsistent conclusion that $2^\lambda < 2^\lambda$ but why doesn't this strict inequality apply in the second step of the equation above?
 A: I don't see why these two facts should be in tension. They have fundamentally different "shapes:"


*

*Cantor's theorem compares base to power: it might help to rewrite it as $$\kappa^2<2^\kappa.$$

*The fact that $\rho^\kappa$ can equal $\lambda^\kappa$ even if $\rho<\lambda$ compares base to base.
So while both these facts tell us something about exponentiation, they don't seem relevant to each other. I think the issue may be that Cantor's theorem agrees with the finite situation while the latter result doesn't, but that shouldn't be too surprising - there's no reason for the behavior of infinite sets to always resemble or always contrast with the behavior of finite sets.
A: Since $\kappa\lt2^\kappa$, $\kappa^\lambda\le2^{\kappa\lambda}=2^\lambda$. You've proven this second inequality isn't strict. There's only an inconsistency if you can prove $\kappa\lt2^\kappa$ would imply $\kappa^\lambda\lt2^{\kappa\lambda}$ as well. If you can prove this, let us know; it'd be exciting to know ZFC is inconsistent. In the mean time, we have a proof that $\kappa^\lambda\not\lt2^{\kappa\lambda}$, so let's just accept that.
