<s>[INST] Series representation of the reciprocal of prime In the article on unique primes in Wikipedia. It is stated that the representation of the reciprocal of a prime $p$ in the numeral base $b$ is periodic of period $n$ if
$$\displaystyle {\frac {1}{p}}=\sum _{i=1}^{\infty }{\frac {q}{(b^{n})^{i}}}$$ where $q$ is a positive integer smaller than $b^n$
I have no idea how this particular representation is obtained. Kindly provide a proof or point to some resources where I can find this representation of $\frac{1}{p}$.
Thanks in advance !
Update:
Let $p$ be a prime and $n$ be the period of $\displaystyle \frac{1}{p}$ in base $b$ and $b \not\mid p$.
If,  $\displaystyle \frac{1}{p} = (0.\overline{a_1a_2a_3\dots a_n})_b$ where $a_1, a_2, a_3, \dots \in \mathbb{Z}_b$
Then, $\displaystyle \frac{1}{p} = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right) = \sum_{i=1}^{\infty}\frac{q}{(b^n)^i}$, where $ q = a_1a_2a_3 \dots a_n$.
I am not sure how $(0.\overline{a_1a_2a_3\dots a_n})_b = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right)$ holds
 [/INST] 
I am not sure how $(0.\overline{a_1a_2a_3\dots a_n})_b = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right)$ holds

Well, $(0.\overline{a_1a_2a_3\dots a_n})_b=0.a_1a_2a_3\dots a_na_1a_2a_3\dots a_na_1a_2a_3\dots a_n\dots$
$=\dfrac{a_1}{b}+\dfrac{a_2}{b^2}+\cdots\dfrac{a_n}{b^n}+\dfrac{a_1}{b^{n+1}}+\dfrac{a_2}{b^{n+2}}+\cdots\dfrac{a_n}{b^{2n}}+\dfrac{a_1}{b^{2n+1}}+\dfrac{a_2}{b^{2n+2}}+\cdots+\dfrac{a_n}{b^{3n}}+\cdots$
$=\dfrac{a_1b^{n-1}+a_2b^{n-2}\cdots+a_n}{b^n}+\dfrac{a_1b^{n-1}+a_2b^{n-2}+\cdots+a_n}{b^{2n}}+\dfrac{a_1b^{n-1}+a_2b^{n-2}+\cdots+a_n}{b^{3n}}+\cdots$
$=a_1a_2\dots a_n\left(\dfrac1{b^n}+\dfrac1{b^{2n}}+\dfrac1{b^{3n}}+\cdots\right).$
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