Motivate why $a^{-n}$ equals to $\frac{1}{a^n}$ I have to prove that $a^{-n} = \frac{1}{a^n}$ with $\frac{3^4}{3^7}$, but before I can do that I have to understand the background. 
The background says:
we know that $\frac{3^4}{3^7} = \frac{1}{3^3}$ and that $3^{4-7} = 3^{-3} = \frac{1}{3^3}$  
I don't understand why $\frac{3^4}{3^7} = \frac{1}{3^3}$, before I can move on and solve why $a^{-n}$  equals to  $\frac{1}{a^n}$.

If $a^b \cdot a^c = a^{b + c}$ generalizes from positive $b, c$ to all integer values of $b, c$, then we have
$$1 = a^0 = a^{1 - 1} = a^1 \cdot a^{-1} = a \cdot a^{-1}.$$
That is, $$a \cdot a^{-1} = 1$$ Hence we are motivated to now define 
$$a^{-1} = \frac{1}{a}$$
Similarly,
$$a^n \cdot a^{-n} = 1$$
and hence we also define
$$a^{-n} = \frac{1}{a^n}$$
