Cardinality: does $|A^B|=|2^B|$ if $|A| \le |2^B|$? Taking finite powers of a countably infinite set still yields a countable set:
$$|\mathbb{N}|=|\mathbb{N}^k|.$$
It's also known that countable powers of the continuum still have the same cardinality as the continuum:
$$|2^\mathbb{N}|=|\mathbb{R}|=|\mathbb{R}^k|=|\mathbb{R}^\mathbb{N}|.$$
Furthermore, taking the next highest ordinal to the power of the continuum doesn't seem to change it. (at least according to wikipedia) Denote $X:=2^\mathbb{R}$, then
$$|2^\mathbb{R}|=|X|=|X^k|=|X^\mathbb{N}|=|X^\mathbb{R}|$$
See a pattern? 
It seems that when taking one set to the power of another set, it doesn't really matter how big the exponent is so long as it is smaller than the base. Or thinking about it in terms of the other variable, if the exponent isn't too big then you might as well replace the base by 2. Ie.,
$$|A| \le |2^B| \Rightarrow |A^B|=|2^B|.$$
Is there a way in which this pattern can be stated and proved rigorously, or is it just a coincidence?

If $B$ is infinite, then yes: $|A^B|\leq |(2^B)^B|=|2^{B\times B}|=|2^B|$ while the other direction is trivial.
