Why does dividing a number with $n$ digits by $n$ $9$'s lead to repeated decimals? For example, $\frac{1563}{9999} = 0.\overline{1563}$.
Why does that make sense from the way the number system works? 
I can vaguely see that since the number $b$ with $n$ $9$'s is always greater than the dividend $a$ of $n$ digits (except in the corner case where they are the same), you will always get $0$ for the units place, after which in the long division process you will be able to divide $a \cdot 10^n$ by $10^n$, $a$ times, which implies you'll be able to divide $a \cdot 10^n$ by $b$, $a$ times with a remainder of $a$, thereby perpetuating this process over and over.
However, I feel this isn't a very precise way to think about it...

This happens because that number $10^n-1=$bunch of $9$'s (for $n$ a natural number) is not divisible by only $2$ and $5$. Any denominator with prime factors other than $2$ and $5$ will have a repeating decimal (it maybe fairly long).
