I need to prove, without assuming the Axiom of Choice, that for cardinals $a,b,b'$, if $a\ge 2$ and $b<b'$, then $a^b <a^{b'}$.
I have already proved that for cardinals $c,c',d,d'$, if $c\neq0$, $c\leq c'$ and $d\leq d'$, then $c^d\leq c'^{d'}$. So I think I need to assume for a contradiction that $a^b=a^{b'}$ and find a contradiction.
Of course, I couldn't come up with a counterexample, so I guess it is true.