<s>[INST] Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$? I was doing an exercise on exponents:
$$\begin{align}
\left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\ 
&= 3^{16} \times 7^{-6} \\
&= \frac{3^{16}} {7^{6}} \\
\end{align}$$
Why did $7^{-6}$ turn to $7^{6}$? More generally, why does a negative exponent turn positive when moved to the denominator?  Would appreciate kindergarten language ;-D
 [/INST] The OP asked for kindergarten language. At that grade level math rules are stressed. So if you are taking algebra-precalculus and this is your last math course, here are a couple of rules;
$\tag 1 \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} \;\text{ where } b \ne 0 \text{, } d \ne 0$
$\tag 2 x = \frac{x}{1}$
$\tag 3 x^m x^n = x^{m+n} \;\text{ where } x \gt 0 \text{ and both } m \text{ and } n \text{ are integers}$
$\tag 4 x \times 1 = x$
$\tag 5 \frac{x}{x} = 1 \;\text{ where } x \ne 0$
$\tag 6 x^0 = 1 \;\text{ where } x \gt 0$
$\tag 7 (-x) + (+x) = 0  \;\text{ for any } x$
So if for some reason your instructor wants you to get rid of negative exponents, and you see $3^{16} \times 7^{-6}$, you can use these seven rules:
$3^{16} \times 7^{-6} = \frac{3^{16} \times 7^{-6}}{1} \; \text{ by (2)}$ 
$\frac{3^{16} \times 7^{-6}}{1} = \frac{3^{16} \times 7^{-6}}{1} \times 1 \; \text{ by (4)}$ 
$\frac{3^{16} \times 7^{-6}}{1} \times 1 = \frac{3^{16} \times 7^{-6}}{1} \times \frac{7^6}{7^6}  \; \text{ by (5)}$ 
$\frac{3^{16} \times 7^{-6}}{1} \times \frac{7^6}{7^6} = \frac{3^{16} \times 7^{-6} \times 7^{+6}} {1 \times 7^{+6} }  \; \text{ by (1)}$ 
$\frac{3^{16} \times (7^{-6} \times 7^{+6})} {1 \times 7^{+6} } = \frac{3^{16} \times (7^{-6 + +6}) } {7^{+6} }  \; \text{ by (3) & (4)}$ 
$\frac{3^{16} \times (7^{(-6) + (+6)}) } {7^{+6} } = \frac{3^{16} \times (7^{0})} {7^{+6}} = \frac{3^{16} \times 1} {7^{+6}} \; \text{ by (7) & (6)}$ 
$\frac{3^{16} \times 1} {7^{+6}} = \frac{3^{16}} {7^{6}}\; \text{ by (4)}$ 

You now have a new rule that you can use:
$\tag 8 x \times y^{-n} = \frac{x}{y^n}  \;\text{ where } y \gt 0 \text{ and } n \text{ is any integer}$
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