Proof for $\frac{x^n}{x^n}=x^{n-n}$ Does the equality $\forall n,x\not=0: \frac{x^n}{x^n}=x^{n-n}$ come straight from the definition of exponents, or is a more elaborate proof needed?
Please note that I'm not asking about $x^0=1$, but only about the $\frac{x^n}{x^n}=x^{n-n}$ part of the equation.
 A: $$\large \dfrac{x^n}{x^n}= x^n \cdot \frac{1}{x^n} = x^n \cdot x^{-n} = x^{(n\ + \ - n)} = x^{(n\ - \ n)} = (x^0 = 1).$$
Note: just as $\dfrac{1}{a}$ is the multiplicative inverse of $a$, (and can be represented by $a^{-1}$). The multiplicative inverse of $a$ is the number you need to multiply by to arrive at $1$: That is, we need $a^{-1}$ to be such that $a\cdot a^{-1} = a^{-1} \cdot a = 1$ (since $1$ is the multiplicative identity for the real numbers: any number multiplied by $1$ remains unchanged). This works for all non-zero $a$ provided $a^{-1} = \dfrac1a.$ Then we have that $a \cdot \dfrac1a = 1$ $$\large \dfrac{1}{x^n} = \underbrace{\frac1x\cdot \frac1x \cdots \frac1x}_{n-times} = \underbrace{x^{-1}\cdot x^{-1} \cdot \cdots x^{-1}}_{n-times} = x^{\overbrace{-1 + -1 + \cdots + -1}^{n - times}} = x^{-n}$$ 
So I'd say you can prove this simply by using the exponent rules.
You might also want to do so using induction, but might be incorporating the use of rules of exponentiation, implicitly. It's sort of a "which came first, the chicken or the egg" question as to which definition is most primitive.
A: Assuming $x \neq 0$,
$$\frac{x^n}{x^n} = 1 = x^0 = x^{n-n}.$$
The only exponent rule used in the above is $x^0 = 1$.  Note, this is a special case of the exponent rule
$$\frac{x^a}{x^b} = x^{a-b}.$$
So, if you already knew that rule, then it follows immediately by letting $a = b = n$.
A: Base Case:
Let n = 0 
$\frac{x^{0}}{ x^{0}} = \frac{1}{1} =  1$
$x^{n-n}  = x^{0 – 0} = x^{0} = 1$
So the base case checks out.
Inductive Hypothesis: 
Want to show that  $\frac{x^{n+1}}{x^{n+1}} = x^{(n+1) – (n+1)}$
$\frac{x^{n+1}}{x^{n+1}} = \frac{x^{n} \cdot x}{x^{n} \cdot x} = x^{n-n} \cdot x^{1-1} =  x^{(n+1) – (n+1) } $
