Hint$\ $ Consider what it means for a real $\rm\ 0 < \alpha < 1\ $ to have a periodic decimal expansion:
$$\begin{array}{rrl} &\alpha\!\!\! &\rm =\,\ 0\:.\overbrace{a_1a_2\cdots a_n^{\phantom{l}}}^{\textstyle a}\ \overbrace{\overline{c_1c_2\cdots c_k}}^{\textstyle \overline c}\ \ \text{ in radix } 10\\[.2em]
\iff& \beta \!\!&\!:= \rm 10^n\, \alpha\! -\! a = 0\:.\overline{c_1c_2\cdots c_k}\\[.2em]
\iff& 10^k\: \beta \!\!&=\ \rm c + \beta\\[.2em]
\iff& (10^k-1)\ \beta \!\!&=\ \rm c\\[.2em]
\iff&\!\!\!\! \rm (10^k-1)\ 10^n\: \alpha \!\!&\in\ \Bbb Z
\end{array}\qquad$$
Thus to show that a rational $\rm\,\alpha\,$ has such a periodic expansion, it suffices to find $\rm\,k,n\,$ as above, i.e. so that $\rm\,(10^k-1)\,10^n\,$ serves as a denominator for $\rm\,\alpha.\,$ Put $\rm\,\alpha\, = a/b,\,$ and $\rm\, b = 2^i\,5^j\,d,\,$ where $\rm\,2,5\,\nmid d\,.\,$ Choosing $\rm\,n\, >\, i,\,j\,$ ensures that $\rm\,10^n\,\alpha\,$ has no factors of $\rm\,2\,$ or $\rm\,5\,$ in its denominator. Hence it remains to find some $\rm\,k\,$ such that $\rm\,10^k-1\,$ will cancel the remaining factor of $\rm\,d\,$ in the denominator, i.e. such that $\rm\,d\,|\,10^k-1\,,\,$ or $\rm\,10^k\equiv 1\pmod{d}\,.\,$ Since $\rm\,10\,$ is coprime to $\rm\,d,\,$ by the Euler-Fermat theorem we may choose $\rm\,k = \phi(d),\,$ which completes the proof sketch. For the converse, see this answer.
Best Answer
Suppose that the decimal is $x=a.d_1d_2\ldots d_m\overline{d_{m+1}\dots d_{m+p}}$, where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then
$$10^mx=10^ma+d_1d_2\dots d_m.\overline{d_{m+1}\dots d_{m+p}}\;,\tag{1}$$ and
$$10^{m+p}x=10^{m+p}a+d_1d_2\dots d_md_{m+1}\dots d_{m+p}.\overline{d_{m+1}\dots d_{m+p}}\tag{2}\;.$$
Subtract $(1)$ from $(2)$:
$$10^{m+p}x-10^mx=(10^{m+p}a+d_1d_2\dots d_md_{m+1}\dots d_{m+p})-(10^ma+d_1d_2\dots d_m)\;.\tag{3}$$
The righthand side of $(3)$ is the difference of two integers, so it’s an integer; call it $N$. The lefthand side is $\left(10^{m+p}-10^m\right)x$, so
$$x=\frac{N}{10^{m+p}-10^m}=\frac{N}{10^m(10^p-1)}\;,$$
a quotient of two integers.
Example: $x=2.34\overline{567}$. Then $100x=234.\overline{567}$ and $100000x=234567.\overline{567}$, so
$$99900x=100000x-100x=234567-234=234333\;,$$ and
$$x=\frac{234333}{99900}=\frac{26037}{11100}\;.$$