# 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.