Unique way to express a positive rational number In the book of G.H. Hardy, "A course of pure mathematics", I found this miscellaneous problem from chapter I, is the exercise 2:

Prove:

  
*Any positive rational number can be expressed in one and only one way in the form
  $$
a_{1} + \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \cdots + \frac{a_{k}}{k!}
$$
  where $a_{1},a_{2},a_{3},\cdots,a_{k}$ are integers, and
  $$
0 \leq a_{1},0 \leq a_{2} < 2,0 \leq a_{3} < 3, \cdots , 0 < a_{k} < k 
$$

I have left this image for reference, which is from the exercise in the book: 

I have some ideas how to prove this, maybe using the representation of $e^{x}$ as an infinite sum, but in general, I have no idea how to solve this problem.
 A: Sketch: Let $\frac{p}{q}$ be the rational number that is being calculated. 


*

*Since $a_i<i$ for $i>1$, we know that 
$$
\sum_{i=k}^m\frac{a_i}{i!}\leq\sum_{i=k}^m\frac{i-1}{i!}=\sum_{i=k}^m\frac{1}{(i-1)!}-\sum_{i=k}^m\frac{1}{i!}=\sum_{i=k-1}^{m-1}\frac{1}{i!}-\sum_{i=k}^m\frac{1}{i!}=\frac{1}{(k-1)!}-\frac{1}{m!}<\frac{1}{(k-1)!}.
$$

*We define each $a_i$ iteratively.  In particular, we choose $a_k$ so that $0\leq\frac{p}{q}-\sum_{i=1}^k\frac{a_k}{k!}<\frac{1}{k!}$.  If $a_k$ is too small, then the remainder is too large and could never be the sum of the remaining terms and if $a_k$ is too large, then the sum is larger than $\frac{p}{q}$.  This shows uniqueness (but not existence - this procedure might never finish).

*To show existence, we could use induction.  We'll show that 
$$
\frac{p}{q}=\sum_{i=1}^q\frac{a_i}{i!}.
$$
We write
$$
\frac{p}{q}=\frac{p(q-1)!}{q!}.
$$
Define $p_q=p(q-1)!$.  Define $a_q$ so that $p_q\equiv a_q\pmod q$.  Then, 
$$
\frac{p}{q}-\frac{a_q}{q!}=\frac{p_q-a_q}{q!}=\frac{p_{q-1}}{(q-1)!}
$$
for some integer $p_{q-1}$.  We can continue in this way, where $a_i$ is defined so that $p_i\equiv a_i\pmod i$.  Since the denominator is always decreasing, eventually, the result is an integer, which is $a_1$.  This construction shows existence, but not uniqueness.
