For the if direction: we construct a way to get $\frac 1k$ for all $k <p$. To start with, we can easily get $\frac{1}{p - 1}$. To get $\frac{1}{p - 2}$, notice that $\frac{2}{p - 2} - \frac{2}{p - 1} = \frac{1}{(p-2)(\frac{p - 1}{2})}$, which wins since the denominator is a multiple of $p-2$. To get $\frac{1}{p - 3}$, consider the unique number $0 < x < 3$ with $3 | p -3 + x$, then consider $\frac{3}{p - 3} - \frac{3}{p -3 + x} = \frac{x}{(p-3)(\frac{p-x}{3})}$, which can give us $\frac{x}{p - 3}$. In general, given that we already have $\frac{1}{p - i}$ for $0 < i < k $, we can always reduce $\frac{x}{p -k }$ to $\frac{y}{p - k}$ with $0 < y < x$, with $x | p - k +y$ by considering $\frac{x}{p - k} - \frac{x}{p - k + y}$. Thus we can always reduce $\frac{k}{p -k}$ to $\frac{1}{p-k}$. By induction, we are done. Only if direction: For any prime $p$ dividing $h$, let $x = \max(v_p(i) \forall i \in [1,h))$. Then consider a good fraction $\frac ab$. Assume to the contrary that this fraction has simplified denominator dividing $p^x$. Notice $v_p(b) \ge x \implies v_p(b) = x \implies v_p(a) \ge 1$, so this is impossible.

@ezpotd i think thats right!