4 is a perfect power

$A \times B = -4$ is also true. If $A = -2$ and $B =2$ then $A^B = (-2)^2 = 4,$ so $A \times B = -2 \times 2 = -4.$

This is a general solution for all values of n and not only integers. For an example,if $A = 16, B = \dfrac12$ then ,

$A^B = 16^\frac12 = \sqrt{16} = 4$

Now, $A \times B = 16 \times \dfrac12 = 8 \ne 4$

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On generalizing, If $A = 4^n, B = \dfrac1n, (n >1)$ then : $\large (4^n)^\frac1n = \sqrt[n]{4^n} = 4$ But, $A \times B = \dfrac{4^n}{n} \ne 4$