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

The question has thoroughly been answered but I do want to contribute with this table that I like.
Begin with this table:
\begin{array}{ccl}
7^5 & = & 7\cdot 7\cdot 7\cdot 7\cdot 7 
\\
7^4 & = & 7\cdot 7\cdot 7\cdot 7
\\
7^3 & = & 7\cdot 7\cdot 7
\\
7^2 & = & 7\cdot 7 
\\
7^1 & = & 7
\end{array}
Going up:


*

*The previous line is multiplied by $7$ on the right hand side.

*The exponent is increased by $1$ on the left hand side.


Going down:


*

*The previous line is divided by $7$ on the right hand side.

*The exponent is decreased by $1$ on the left hand side.


What would be a natural extension? To keep doing this pattern both up and down! And thus:
\begin{array}{ccl}
7^5 & = & 7\cdot 7\cdot 7\cdot 7\cdot 7 \phantom{\dfrac1{7^3}}
\\
7^4 & = & 7\cdot 7\cdot 7\cdot 7\phantom{\dfrac1{7^3}}
\\
7^3 & = & 7\cdot 7\cdot 7\phantom{\dfrac1{7^3}}
\\
7^2 & = & 7\cdot 7 \phantom{\dfrac1{7^3}}
\\
7^1 & = & 7\phantom{\dfrac1{7^3}}
\\
7^0 & = & \dfrac77 = 1\phantom{\dfrac1{7^3}}
\\
7^{-1} & = & \dfrac17 \phantom{\dfrac1{7^3}}
\\
7^{-2} & = & \dfrac{1/7}7 =  \dfrac1{7^2}
\\
7^{-3} & = & \dfrac{1/7^2}7 = \dfrac1{7^3}
\end{array}
