This question seems to show up a lot. Given cardinal numbers $\kappa$ and $\nu$, under what circumstances do we have $\kappa^\nu = 2^\nu$? Assume $\nu$ is infinite.
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Clearly $\kappa^\nu\le1$ if $\kappa\le1$.
Suppose $2\le\kappa$. Then $2^\nu\le \kappa^\nu$. Suppose now that $\kappa\le2^\nu$. Then $\kappa^\nu\le(2^\nu)^\nu=2^\nu$, so we have that if $$2\le \kappa\le 2^\nu,$$ then we have equality (by the Bernstein-Schroeder theorem).
Of course, this is best possible: If $\kappa\ge(2^\nu)^+$, then $\kappa^\nu\ge((2^\nu)^+)^\nu\ge(2^\nu)^+>2^\nu$.
Note that I am using freely the axiom of choice. For example, in the equality $(2^\nu)^\nu=2^\nu$, which follows from the straightforward $(2^\nu)^\nu=2^{\nu\cdot\nu}$, and the equality $\nu\cdot\nu=\nu$.
See also this question.
Andrés E. Caicedo
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