Why does the Numerator of the Denominator go to the Denominator?

Question

$\left(\dfrac 25\right)^{-3} = \dfrac{1}{\left(\frac 25\right)^3} = \dfrac{1}{\frac 8{125}} = \dfrac {125}8$

I'm not sure if my intuition about the above is correct, but here goes:
My intuition tells me that since we are taking the reciprocal of a fraction, the numerator of the "higher fraction" ($1$) is the whole of the denominator of the "lower fraction" ($125$), and so we can move the denominator of the lower fraction as the numerator of the higher fraction, because $1$ is practically the same as $125$.

What I can't seem to grasp is then why does the numerator of the lower fraction ($8$) become the denominator of the new fraction?

My question is very much like this one, but rotated.

Answer 1

So you don't understand this part:

$$\dfrac{1}{\frac 8{125}}=1 \div\dfrac 8{125}=1\times \dfrac{125}8=\dfrac {125}{8}$$

Can you understand it from there?

Answer 2

By definition $\frac{1}{a}$ is the inverse of $a$, i.e. the number such that $a \cdot \frac{1}{a}=1$.

So, what is the inverse of $\frac{8}{125}$?

Answer 3

I managed to make sense of this one by looking at the fractions from a decimal point of view.

$\dfrac {1}{\frac {8}{125}} = \dfrac {1}{0.064}$

$1$ is $15.625$ times larger than $0.064$. So that also means that $125$ is $15.625$ times larger than $8$. Thus,

$\dfrac {1}{\frac {8}{125}} = \dfrac {1}{0.064} = \dfrac {125}{8}$

While I appreciate the other guys' answers, (and some will probably feel they explained it sufficiently) I found the above to give the "intuitive leap" to better understanding why the bottom fraction is flipped up, for someone not well versed in mathematics.