Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$?
Notice that $7^6\cdot 7^{-6}=7^{6-6}=7^0=1$
Notice also that $7^6\cdot\frac{1}{7^6}=\frac{7^6}{7^6}=1$
So, we learned that $7^6\cdot 7^{-6}=7^6\cdot\frac{1}{7^6}$.
Remembering that $x\cdot a=x\cdot b$ for nonzero $x$ implies that $a=b$ by cancelling this tells us that $7^{-6}=\frac{1}{7^6}$
In general the following properties are useful to know:
- $x = \frac{x}{1}=x^1$
- $x^n = \frac{1}{x^{-n}}$
- $x^{-n}=\frac{1}{x^n}$
Another useful identity is $x^0 = 1$ which is true for all nonzero $x$
Tangentially, depending on context it can also be correct to say that $x^0=1$ for $x=0$ as well, for example in the field of combinatorics. There are some other situations though where we leave $0^0$ undefined.
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 $7$: $$ \dots \qquad7^{-2} = \frac{1}{49},\qquad 7^{-1} = \frac{1}{7},\qquad7^0=1,\qquad7^1=7,\qquad 7^2=49,\qquad\dots $$
Exponentiation is initially understood as repeated multiplication with the exponent indicating how many times the base appears as a factor in the product. Thus, for example, $2^3=2\times2\times2$ since $2$ appears as a factor three times.
Given this initial understanding of exponentiation certain properties can be observed such as $x^2\cdot x^3=(x\cdot x)\cdot(x\cdot x\cdot x)=x^5$. From examples such as this we discover that
$$ x^m\cdot x^n=x^{m+n} \tag{1}$$
Similarly, it can be seen that $(x^3)^2=x^3\cdot x^3=x^6$, for example. From such examples we discover that
$$(x^m)^n=x^{mn} \tag{2}$$
But these rules assume that $n$ is a positive integer. What if that is not the case? Can meaning be given to exponentiation in such a way that rules $(1)$ and $(2)$ are preserved?
Let us see, for example, if meaning can be given to exponentiation with $0$.
We would want it to be true that $x^0\cdot x^n=x^{0+n}=x^n$ in order to satisfy rule ${1}$. But that means that it would have to be the case that $x^0=1$. So that is how exponentiation to the power $0$ is defined.
But then what about exponentiation to a negative integer power? How could sense be given to a term like $x^{-n}$?
Well, in order for rule $(1)$ to apply it would have to be the case that
$$ x^{-n}\cdot x^n=x^{-n+n}=x^0=1 $$
But if $x^{-n}\cdot x^n=1$ then it must be the case that $x^{-n}=\dfrac{1}{x^n}$ and $x^n=\dfrac{1}{x^{-n}}$.