Prove that every positive rational number can be expressed as a quotient of products of numbers of the form ${n\choose \lfloor n/2\rfloor}$
Below is my attempt. I'm not sure how to express fractions such as $(2k)/(2k+1)$ in the desired form.
Let $a/b$ be a positive rational number. Note that we can swap the numerator and denominator in the quotient of products of numbers of the form ${n\choose \lfloor n/2\rfloor}$ that equals $a/b$ to write $b/a$ in the desired form (*). By this fact and the fact that $1 = {1\choose \lfloor 1/2\rfloor}$, we may assume WLOG that $a<b$. Observe that $a/b = a/(a+1)\cdot (a+1)/(a+2)\cdots (b-1)/b,$ so it suffices to prove the claim for fractions of the form $a/(a+1)$. We first deal with the case of fractions of the form $(2k+1)/(2k+2)$. Note that ${2k+1\choose k}/{2k+2\choose k+1} = (2k+1)!/(k!(k+1)!)\cdot (k+1)!^2/(2k+2)! = (k+1)/(2k+2) = 1/2$ and that ${2k\choose k}/{2k+1\choose k} = (2k)!/(k!)^2 \cdot (k!(k+1)!)/(2k+1)! = (k+1)/(2k+1).$ Then ${2k+1\choose k}/{2k\choose k}\cdot {2k+1\choose k}/{2k+2\choose k+1} = (2k+1)/(2k+2),$ as desired. Now we deal with the case of fractions of the form $(2k)/(2k+1).$ I'm not sure how to deal with this case. As a specific example, one can try expressing $6/7$ in the desired form. I get $6 = {3\choose 2}/{2\choose 1}\cdot {4\choose 2}/{3\choose 2}\cdot {6\choose 3}/{5\choose 2}$, but I'm not sure how to get $1/7$.
It may be useful to consider cases where n is prime.
OP has everything that they need, and just need to push through the algebra.
OP's idea of chaining together $\frac{k+1}{k}$ terms to get $ \frac{a}{b}$ is a good idea, but hard to execute on. It's easier to get $a$ directly.