$\large \dfrac{x^n}{x^n}= x^n \cdot \frac{1}{x^n} = x^n \cdot x^{-n} = x^{(n\ + \ - n)} = x^{(n\ - \ n)} = (x^0 = 1).$
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.