Period length in the decimal expansion of fractions \frac{1}{n} \frac{1}{7} is

Step 1
For ' p co' to 10 you have by Fermat's little theorem that
${\displaystyle p\rceil \sim [{\left(10\right)}^p-1].}$
This means that there exists an integer of the form
${\displaystyle d_{p-1}{\left(10\right)}^{p-1}+d_{p-2}{\left(10\right)}^{p-2}+\cdots +d_0{\left(10\right)}^0}$
where
${\displaystyle d_{p-1},d_{p-2},\cdots d_0\in \{0,1,\cdots ,9\}}$
such that
${\displaystyle \frac{1}{p}=\frac{d_{p-1}{\left(10\right)}^{p-1}+d_{p-2}{\left(10\right)}^{p-2}+\cdots +d_0{\left(10\right)}^0}{{\left(10\right)}^p-1}.}$
Further, any fraction of the form
${\displaystyle \frac{d_{p-1}{\left(10\right)}^{p-1}+d_{p-2}{\left(10\right)}^{p-2}+\cdots +d_0{\left(10\right)}^0}{{\left(10\right)}^p-1}}$
will have a decimal representation of
${\displaystyle 0.\overline{d_{p-1}\sim d_{p-2}\sim \cdots \sim d_0}.}$
Therefore, for any ' (p) co' to (10), the (infinite) decimal representation of ${\displaystyle \left(\frac{1}{p}\right)}$ will either have a period of ${\displaystyle (p-1)}$, or a period of k, where k divides ${\displaystyle (p-1)}$
The only time that the period will be $k<(p-1)$ is if p happens to divide ${\displaystyle [{\left(10\right)}^k-1]}$
(7) (for example) is not a divisor of either (99) or (999). However, (13) which you know has to be a divisor of ${\displaystyle [{\left(10\right)}^{\left(12\right)}-1]}$ also happens to be a divisor of ${\displaystyle ({10}^3+1)}$
This implies that (13) is a divisor of
${\displaystyle ({10}^3+1)({10}^3-1)=({10}^6-1).}$
This explains why the decimal representation of ${\displaystyle \frac{1}{13}}$ has a period of (6), rather than (12).
Step 2
An example of an unusual way of using the above analysis is to use it to indirectly conclude that the fraction ${\displaystyle \left(\frac{1}{11}\right)}$ has a period of (2).
This can be reasoned directly simply by noting that ${\displaystyle 11\mid 99}$
The (convoluted) indirect way is to notice that the period of ${\displaystyle \left(\frac{1}{11}\right)}$ must either be (10) or a divisor k of (10).
Further, you know that ${\displaystyle 11\rceil \sim [{\left(10\right)}^3+1].}$
This implies that ${\displaystyle 11\rceil \sim [{\left(10\right)}^6-1].}$
This means that the period k of ${\displaystyle \left(\frac{1}{11}\right)}$ must be a common divisor of both (6) and (10). This allows you to indirectly conclude that ${\displaystyle \left(k\right)=2}$.