I was following a proof, the crux of which came down to understanding that
$$\ \frac{10^k}{10^h}\equiv \frac{-1}{1}\ \mod(p)$$ for some $\ h < k $ and prime $p$, provided that
$\ {10^k}\equiv {-1}\ \mod(p)$ and $\ {10^h}\equiv {1}\ \mod(p)$ are already known to be true.
I know that there are very specific rules for modular division (although I am ignorant to them), so I tried out another example
$$\ \frac{4^3}{4^2}\equiv \frac{-1}{1}\ \mod(5)$$
in which, sure enough, the "division" turned out to be true. Please help me out, what the heck is this sorcery?
If $10^h\equiv1\pmod p$ and $10^k\equiv-1\pmod p$ where $h<k$ then $$-1\equiv 10^k=(10^h)(10^{k-h})\equiv(1)(10^{k-h})=10^{k-h} \pmod p.$$ Thus $10^{k-h}\equiv-1\pmod p$.
Note that I have used standard facts about modular multiplication here, but nothing about "division".