In a test where the topic was cardinal arithmetic, one student claimed that $a^b=a$ for any infinite cardinals such that $a>b$ (and used this claim in some computations). It is clear that $a\le a^b$, but it is possible that the inequality is strict.
Coincidentally, the student got the correct result, since all cardinals in this problem were beth numbers and for such numbers you cannot get a counterexample. (If $a=2^c$ where $c\ge b$, then we get $a^b=2^{c\cdot b} = 2^c = a$.)
The simplest counterexample I was able to come up with was $$\aleph_\omega<\aleph_\omega^{\aleph_0}.$$ Indeed we have $\operatorname{cf}(\aleph_\omega)=\aleph_0$, so we get from König's theorem that $$\aleph_\omega<\aleph_\omega^{\operatorname{cf}(\aleph_\omega)}=\aleph_\omega^{\aleph_0}.$$
Are there some other simple (interesting, insightful) counterexamples to this claim? I.e., some other examples of infinite cardinals such that $$a>b\text{ and }a<a^b.$$ Some comments on an example I suggested - for example, if it can be simplified or generalized - are welcome as well. (To make things easier, let us work in ZFC; i.e., we assume that Axiom of Choice holds.)
Assuming $\sf GCH$, these are the only counterexamples you will find. And since it is consistent with $\sf ZFC$, you can't do much better.
To see this is the case, note that $\sf GCH$ implies that for all regular $\kappa$, $\kappa^{<\kappa}=\kappa$.
If there is some place the countinuum hypothesis fails, it will be the counterexample you want anyway.