Inequality of cardinals: $AB<A^B$ for $B\geq |\mathbb{N}|$

Question

Let $A$ and $B$ be cardinal numbers. Assume $B\geq |\mathbb{N}|$. Is there an easy proof of $$ AB<A^B? $$ (Note the strict inequality!)

Answer 1

No, as equality can be obtained:

Let $A:= 2^\omega$ (continuum) and let $B:= \omega$.
Then $AB=\max(A, B) = A$ and $A^B= 2^{\omega\cdot\omega} = A$.