For cardinals $a,b,b'$, if $a\ge 2$ and $b<b'$, then $a^b <a^{b'}$

I need to prove, without assuming the Axiom of Choice, that for cardinals $$a,b,b'$$, if $$a\ge 2$$ and $$b, then $$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.

• Why do you think this is true? Also, $a'$ does not appear anywhere in your question except in $a\leq a'$. – Asaf Karagila Apr 14 at 9:25
• $2<3$, but $\aleph_0^2=\aleph_0^3$. – TonyK Apr 14 at 9:27
• I couldn't come up with a counterexample. I meant that in the case that $a=a'$, the result I have already proved implies the inequality I need, but weakly; hence I thought to proceed by assuming for a contradiction that $a^b=a^{b'}$. I'll clarify the notation. – Davide Apr 14 at 9:29
• Ooh thanks @TonyK, that was more straightforward than I expected – Davide Apr 14 at 9:34
• @bof: How do you prove this without choice, though? :) – Asaf Karagila Apr 14 at 9:46