Show $\left|B^A\right| \cdot \left|B^A\right| = \left|B^A\right|$

Question

Let $A$, an infinite set such that $\left|A\right|\cdot \left|A\right| = \left|A\right|$ and Let $B$, an arbitrary set. Show $\left|B^A\right| \cdot \left|B^A\right| = \left|B^A\right|$

I'd be glad to get guidance.
Thanks.

My Try:

Can we prove it directly by:

$$\left|B^A\right|\times \left|B^A\right| = \left|B\right|^{\left|A\right|} \times \left|B\right|^{\left|A\right|} = \left|B\right|^{\left| A \right| + \left| A \right|}= \left|B\right|^\left| A \right|$$

Answer 1

As $A$ is infinite, we have $|A|=|A|+|A|$, i.e. there exist a bijection $\phi\colon A\times\{0,1\}\to A$. Then $$f\mapsto\langle a\mapsto f(\phi(a,0)), a\mapsto f(\phi(a,1))\rangle$$ is a bijection(!) $B^A\to B^A\times B^A$