<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] First you need to understand why
$$
7^{-1} = \frac{1}{7}
$$
In the row below, to move one to the right, multiply by $7$:
$$
7^1=7,\qquad 7^2=49,\qquad 7^3=343,\qquad\dots
$$
And consequently, to move to the left, divide by $7$.  So that is how to extend it the other way: keep dividing by $7$:
$$
\dots \qquad7^{-2} = \frac{1}{49},\qquad 7^{-1} = \frac{1}{7},\qquad7^0=1,\qquad7^1=7,\qquad 7^2=49,\qquad\dots
$$
</s>