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

I've already proven that for cardinals $ a, a, b, b $, if $ a neq0 $, $ a leq a & # 39; $ and $ b leq b $then $ a ^ b leq a ^ # $ {$ b}. It remains for me to refute the equality in the title.

Moreover, I am obliged not to assume the axiom of choice.