How is the start of repeating decimals defined? Today I tried to find the period length of the repeating decimals of 8/86 by asking WolframAlpha (yes a somewhat stupid question because it's the same as 1/43). 

Here I found the repeating decimals being 930232558139534883720; isn't it the same as 093023255813953488372? If I try 1/11 it's 90, again starting with a non-zero digit. 
Is this just a convention or is there a deeper truth? (sorry for this maybe trivial question)
 A: Indeed, it is just a convention. We say that a real number $\alpha$, say in the interval $(0,1)$, is ultimately periodic (in base $10$) if there are two finite sequences of digits (usually called words) $U = d_1d_2\dotsc d_n$ and $V = d_{n+1}d_{n+2}\dotsc d_{n+m}$ such that $\alpha = 0.UV^{\omega}$ in decimal notation, where $V^{\omega}$ stands for the concatenation of $V$ with itself countably many times. The words $U$ and $V$ are called the $preperiod$ and $period$ of $\alpha$, respectively.
As you noticed, if a number is ultimately periodic there are infinitely many choices for $U$ and $V$, and we are free to choose the one that best fits our purposes at a given time. This usually means a representation where the lengths of both $U$ and $V$ are minimal.
I have no clue why Wolfram Alpha gives that particular representation, though. A possible explanation is that internally it represents the period as an integer, so it would lose any leading zeros unless they were included in the preperiod.
A: It is probably a preferred form which the WA computational engine wants to keep its numbers in.
An internal normalization.
This way you have a decimal representation with maximal fixed width $f$
$$
x = 
(d_m\cdots d_0.\underbrace{d_{-1}\cdots d_{-f}}_f\underbrace{\overline{d_{-(f+1)}\cdots d_{-(f+p)}}}_p)_{10} \\
$$
and the periodical part will show up in calculations as
\begin{align}
\sum_{k=1}^\infty \frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^{f+kp}}
&=
\frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^f} 
\sum_{k=1}^\infty \left(\frac{1}{10^p}\right)^k
\\
&=
\frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^f}
\left(\frac{1}{1-1/10^p} - 1\right)
\\
&=
\frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^f(10^p-1)}
\end{align}
It might be convenient to have the first digit of the periodical part $d_{-f+1}$ non-zero.
