What numbers can be created by repetitive division of subsequent natural numbers? I currently don't have the knowledge to write the question down in generalized form. So I will instead give some examples to show what I mean.
$$A = \frac{1}{\cfrac{2}{\left(\cfrac{3}{4}\right)}}$$
The above equation is an example of a 'tower equation'. There are two rules for such equations:


*

*It can be as tall as you want but the floors should always be $1$ to $n$, where the highest floor is $1$ and the ground floor is $n$. Floors are named with natural numbers so $n$ must also be a natural number.

*The order in which you do the divisions is left up to you.
$$B = \cfrac{\cfrac{1}{\left(\cfrac{2}{3}\right)}}{4},C = \cfrac{\left(\cfrac{1}{2}\right)}{\left(\cfrac{3}{4}\right)}, D = \cfrac{\cfrac{\left(\cfrac{1}{2}\right)}{3}}{4}$$ are all valid 'tower equations' with different results.


I've already figured out that a 'tower equation' can come infinitely close to $0$ and it can get higher then 2 as shown by TheSimpliFire (1/((2/3)/4) = 6). But is it possible to construct all fractions using a 'tower equation'?
And, by extension, is it able to create all rational numbers?
 A: The answer is no you cannot get every rational number. For example you cannot get the number $2$. I'll need to use Bertrand's postulate to explain why:

For every $n>1$ there must be a prime $p$ such that $n < p < 2n$. Or written slightly differently, for any $n > 2$ there must be a prime $p$ such that $\lfloor n/2 \rfloor < p < n$.

Suppose you have a tower equation of size $n$. You can resolve all the parenthesis in your tower equation and write it as a proper fraction
$$\frac{k_1 \dotsb k_\ell}{k_{\ell+1} \dotsb k_n}\,.$$
where the $k_i \in \{1, \dotsc, n\}$ and $k_i = k_j \implies i = j$, so the $k_i$ are just some ordering of $\{1, \dotsc, n\}$. One of these values of $k_i$ must be a prime $p$ between $\lfloor n/2 \rfloor$ and $n$ such that $p$ does not divide any of the other numbers $\{1, \dotsc, n\}$ besides itself. So in your proper faction, there will be exactly one factor of $p$ in the numerator or denominator. If your size  $n$ is at least four, this prime isn't $2$, and since we can write out all the values of the tower equations of size four or less (see TheSimpliFire's answer) and see that the value of $2$ doesn't appear, $2$ cannot appear for a larger size tower because of this "un-cancellable" prime $p \neq 2$ that must appear in either the numerator or denominator.

Here's an unrelated thought I had about this question. I don't immediately see how it can help answer the question, but it does mean that the number of tower equations of size $n$ will be the $n^{\text{th}}$ Catalan number.
Think about building a tower as writing down your large fraction without parenthesis, and then choosing the "top-most" fraction bar, dividing your tower into two smaller towers, the numerator and denominator, and then continuing this process recursively on those two smaller towers. Doing this on a tower of size $n$ builds a full binary tree with $n$ leaves, the numerators corresponding to left branches and the denominators corresponding to right branches, and with the leaves labelled in order from left to right with the numbers $\{1, \dotsc, n\}$. Then the number that the tower equals can be calcated from the tree as 
$$\prod_{i\in\{1, \dotsc, n\}} (-1)^{r_i}i$$
where $r_i$ is the number of "right turns" or "right branches" in the tree you'd have to take to get to the leaf $i$. 
A: Not a complete answer: I will come back when I have time
Let's consider a case-by-case approach. Here, '$\text{Product}$' means the product of every distinct fraction.
$n=1$: $$\color{red}{1}\\\text{Product}=1$$
$n=2$: $$\color{red}{\frac12}\\\text{Product}=\frac12$$
$n=3$: $$(1/2)/3=\color{red}{\frac16}\quad1/(2/3)=\color{red}{\frac32}\\\text{Product}=\frac14$$ 
$n=4$: $$((1/2)/3)/4=\color{red}{\frac1{24}}\quad(1/(2/3))/4=\color{red}{\frac38}\\(1/2)/(3/4)=\color{red}{\frac23}\quad1/((2/3)/4)=\color{red}{6}\quad1/(2/(3/4))=\color{red}{\frac38}\\\text{Product}=\frac1{16}$$
$n=5$: $$(((1/2)/3)/4)/5=\color{red}{\frac1{120}}\quad\cdots$$
Obviously, such 'tower fractions' will contain every fraction of the form $1/k!$ where $k$ is a natural number.
It seems like the general product is $$\text{Product}=\cfrac1{2^{2^{n-1}}}$$
