Why is $\frac{1}{x^{-1}}=x$? I believe I get why $x^{-1}=\frac{1}{x}$ . Because if we would like our exponent rule ($\frac{x^n}{x^m}= x^{m-n}$) to be true for all values of $m$ and $n$ then $x^{-1}$ should be another way to write $\frac{1}{x}$ as [$\frac{x^1}{x^2}=x^{1-2}=x^{-1}=\frac{1}{x}$] even if multiplying a certain number by itself negative number of times makes no sense.
So I can see why $x^{-1}$ is just another way to write $\frac{1}{x}$ but I can't get how we've arrived that $\frac{1}{x^{-1}}$ is equal to $x$; I can't think of any way that will give us a negative power in the doninator in order to proof that $\frac{1}{x^{-1}}$=$x$, we can't subtract from the dominator's power even when we divide. We always end up subtracting from the numerator's power instead, example:

$\frac{1}{x}÷\frac{1}{x^3}=\frac{1}{x}×x^3= \frac{x^3}{x}=x^{3-1}=x^2$

 A: For a nonzero real number $x$, the reciprocal $\frac1x$ is the real number such that when multiplying with $x$, you get $1$. In other words, when $x\ne 0$, $\frac1x$ is the multiplicative inverse of $x$.
For $x\ne 0$, the negative exponential $x^{-1}$ is by definition $\frac1x$, and thus
$$
x\cdot x^{-1}=x^{-1}\cdot x=1
$$
Therefore,
$$
\frac{1}{x^{-1}}=x.
$$
A: $\frac{1}{x^{-1}}=\frac{x}{xx^{-1}}=\frac{x}{1}=x$
A: I'm not really good at explaining stuff, but this guy definitely is.
Answering you question, $x^{-1}$ is equal to $\frac{1}{x}$. 
And $\frac{1}{x^{-1}}$ is equal to $\frac{1}{1}/\frac{1}{x}=\frac{1}{1}*\frac{x}{1}=x$
A: “ I can't think of any way that will give us a negative power in the doninator ”
To put something in the denominator, you must divide by it as taught in elementary school.
$1\div x^{-3}=\frac 1 {x^{-3}}$ Then it is in the denominator.
To reiterate: $1 \div x^{-1}=\frac 1 {x^{-1}}$
Now to prove using this, $1 \div x^{-1}=1\times x$, you times the multiplicative inverse.
