A strange number base
Aaaages ago, @vingaints tweeted:
This is pretty wild. It feels like what the Basis Representation Theorem is for Integers but for Rational Numbers. Hmm  trying to prove it now. Feels like a tough one. Need to work some examples! https://t.co/tgcy8iaXHa pic.twitter.com/tgcy8iaXHa
— Ving Aints (@vingAints) September 18, 2018
In case that’s not showing for you, it says:
Any positive rational number acn be expressed in one and only one way in the form
$a_1 + \frac{a_2}{1\cdot2} + \frac{a_3}{1\cdot 2 \cdot 3} + \dots + \frac{a_k}{1\cdot 2 \cdot 3 \dots k}$,
Where $a_1$, $a_2$, … $a_k$ are integers, and $0 \leq a_1$, $0 \leq a_2 \lt 2$, $0 \leq a_3 \lt 3$, … $0 \lt a_k \lt k$.
It’s one of those tweets where I show it to some mathematical friends and they all go ‘ooo!’. So, how would we prove such a thing?
What we need to prove
There are three things we need to prove:

There is at least one way to represent each number in this format

There is at most one way to represent each number in this format

This unique representation terminates
As Ving notes, we can restrict our analysis to the rationals between 0 and 1, given that $a_1$ can be any positive number.
The nugget (a lemma)
I plan to prove, by induction, that the $(n1)$term sequence beginning with $a_2$ can represent every rational number of the form $\frac{j}{n!}$, with $0 \leq j \lt n!$. (Note that this is slightly different from the target sequence, which requires the last term be nonzero. We’ll get to that.)
Base case: For $n=2$, we can clearly make $\frac{0}{2}$ and $\frac{1}{2}$.
Inductive step: Suppose all numbers of the form $\frac{j}{k!}$ can be made for some integer $k$ and for $0 \leq j \lt k!$. Let $0 \leq J \lt (k+1)!$. Now, $J = (k+1)j + m$, where $j = \floor{\frac{J}{k+1}}$ and $m = j \pmod{k+1}$ so that $0 \leq j \lt k!$ and $0 \leq m \lt k+1$  each $J$ can be written uniquely in this form ((Ving suggests the Euclidean algorithm to prove this; I would probably appeal to another pigeonhole argument.)).
We then have $\frac{J}{(k+1)!} = \frac{j}{k!} + \frac{m}{(k+1)!}$, and since we can make any $\frac{j}{k!}$ from a $(k1)$term series, we can make $\frac{J}{(k+1)!}$ from a $k$term series.
Conclusion: Therefore, the proposition is true for $n=2$. If it is true for $n=k$, it is also true for $n=k+1$, so it is true for $n=2,3,4,…$ as required.
Corollary: How many sequences are there of the form $\{a_2, a_3, \dots a_n\}$? Each $a_i$ is a free choice of an integer such that $0 \le a_i \lt i$, of which there are $i$; therefore, there are $2 \times 3 \times 4 \times \dots n = n!$ possible sequences.
Between these $n!$ sequences, all $n!$ fractions of the form $\frac{j}{n!}$ with $0\le j \lt (n!  1)$ can be made — so by the pigeonhole principle, each sequence corresponds to exactly one fraction.
The proof
To show (1): Any number of the form $\frac{p}{q}$, with $p$ and $q$ integers such that $0\lt p \lt q$, can be written in the form described above in exactly one way ((The case where $p=0$ is trivial.)).
Consider $\frac{(q1)!p}{q!} = \frac{p}{q}$. According to the lemma, this can be written in precisely one way using a $(q1)$term sequence, possibly with trailing zeroes.
Since the sequence has at least one nonzero term, it must have a last nonzero term. Truncating the sequence at this last nonzero term gives a unique ((because the $(q1)$term sequence is unique)) terminating ((because it finishes, duh)) representation for $\frac{p}{q}$.
$\blacksquare$
I can’t say that this is an especially practical system of numbers – although it leads ((I think – I suspect it might need a bit more work)) to a straightforward proof that $e$ is irrational (since, by definition, $e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots$, its representation in this form would be ${ 2; 1, 1, 1, 1, \dots }$, which does not terminate  so it is not rational) – but I rather like it.
Does my proof have holes in? Is there a neater way? Let me know in the comments!
* Many thanks to @vingaints for patiently explaining the holes in my proof! * Edited 20190528 to fix a broken link. Thanks, Barney!