A Cantor series is defined as the series $
\sum\limits_{k=0}^{+\infty}{\dfrac{c_k}{k!}}
$, where $\forall k\in \mathbb{Z}_{+}, 0\leq c_{k+1}\leq k$. It can be proven that 'almost all' real numbers can be uniquely expressed in a Cantor series (the exceptions of which is not key to this question). How to prove the following proposition:
A real number $x\in [0,1)$ is rational if and only if there is some $k_0\in\mathbb{Z}_{+}$ such that in the Cantor series of $x$, $\forall k\geq k_0, c_k=0$.
Best Answer
Use induction to show the nontrivial direction: Let $q$ be rational in $[0,1)$, then $q$ can be written as a Cantor series.
Let $b$ be a positive integer. Use induction: Assume that any fraction $\frac{A}{(b-1)!}$; $A$ a nonnegative integer satisfying $A<(b-1)!$ can be written as a cantor series also of the form $$\frac{A}{(b-1)!}=\sum_{k=0}^{b-1} \frac{c_k}{k!}.$$ We now establish Claim 1:
Claim 1: Any fraction $\frac{A'}{b!}$; $A'$ a nonnegative integer satisfying $A'<b!$, can be written as a cantor series also of the form $\frac{A'}{b!}=\sum_{k=0}^{b} \frac{c'_k}{k!}.$
Then to establish Claim 1: First write $bA''+c_b=A'$, where $c_b = A' \pmod b$ [in particular $c_b$ is a nonegative integer strictly less than $b$]. Then note that $A'-c_b$ is a multiple of $b$ and thus $A''$ is a nonnegative integer. Now note the following: $$\frac{A'}{b!} = \frac{c_b}{b!} + \frac{bA''}{b!}$$ $$= \frac{c_b}{b!} + \frac{A''}{(b-1)!},$$ or in particular $$\frac{A'}{b!} \ = \ \frac{c_b}{b!} +\frac{A''}{(b-1)!}.$$
But then as as the string $\frac{A''}{(b-1)!} =$ $\frac{bA''}{b!} \le $ $\frac{A'}{b!} < 1$ holds, it follows that not only is $A''$ a nonnegative integer, but also, $\frac{A''}{(b-1)!}$ is strictly less than $1$. Thus, by the inductive hypothesis, $\frac{A"}{(b-1)!}$ can be rewritten, for some choice of $c_0,c_1, \ldots, c_{b-1}$ as in the OP: $$\frac{A''}{(b-1)!} \ = \ \sum_{k=0}^{b-1} \frac{c_k}{k!}.$$
But then $$\frac{A'}{b!} = \frac{c_b}{b!} + \frac{A''}{(b-1)!} \ = \ \frac{c_b}{b!}+\sum_{k=0}^{b-1} \frac{c_k}{k!},$$ or in particular, $$\frac{A'}{b!} \ = \ \frac{c_b}{b!}+\sum_{k=0}^{b-1} \frac{c_k}{k!},$$ where, as noted previously, $c_b$ is a nonnegative integer strictly less than $b$, and where $c_0,c_1,\ldots, c_{b-1}$, are as in the OP. Thus, indeed, any fraction $\frac{A'}{b!}$; $A'$ a nonnegative integer satisfying $A'<b!$, can be written as a cantor series also of the form $\frac{A'}{b!}=\sum_{k=0}^{b} \frac{c'_k}{k!}.$ And so Claim 1 follows. $\surd$
Note that Claim 1 suffices to prove the result; let $q$ be any rational number in $[0,1)$, then write $q=\frac{a}{b}$, with $a$ and $b$ nonnegative integer satisfying $a<b$. Then $q$ can be rewritten $q = \frac{A'}{b!}$, with $A'$ an integer less than $b!$; indeed $A'=a(b-1)!$.