If we just consider the largest digit of a $k$ digit number $c_k \cdot k!$ this can express $k+1$ different multiples of $k!$: $0$, $k!$, $2 \cdot k!$, $\ldots$, $k \cdot k!$. (Much like how we can express 0, 1000, 2000, 3000, ..., 9000 in decimal)
So for this type of number system to be able to represent every number we need numbers with fewer digits than $k$ - ones of the form $\sum\limits_{i=1}^{k-1} c_i\cdot i!$ - to be able to express everything between the gaps (much like how every number below 10000 is some thousand + a number from 0 to 999). Specifically it needs to be able to express every number from $0$ up to $k!-1$.
Let us check that the biggest number we can express with $k$ digits
Lemma $k! - 1 = \sum_{i=1}^{k-1} i \cdot i!.$
Example
$ \begin{array}{} && (5 \cdot 5! + 4 \cdot 4! + 3 \cdot 3! + 2 \cdot 2! + 1 \cdot 1!) + 1 \\ &=& 5 \cdot 5! + 4 \cdot 4! + 3 \cdot 3! + 2 \cdot 2! + (1 \cdot 1! + 1 \cdot 1!) \\ &=& 5 \cdot 5! + 4 \cdot 4! + 3 \cdot 3! + 2 \cdot 2! + (2 \cdot 1!) \\ &=& 5 \cdot 5! + 4 \cdot 4! + 3 \cdot 3! + 2 \cdot 2! + (1 \cdot 2!) \\ &=& 5 \cdot 5! + 4 \cdot 4! + 3 \cdot 3! + 3 \cdot 2! \\ &=& 5 \cdot 5! + 4 \cdot 4! + 3 \cdot 3! + 1 \cdot 3! \\ &=& 5 \cdot 5! + 4 \cdot 4! + 4 \cdot 3! \\ &=& 5 \cdot 5! + 4 \cdot 4! + 1 \cdot 4! \\ &=& 5 \cdot 5! + 5 \cdot 4! \\ &=& 5 \cdot 5! + 1 \cdot 5! \\ &=& 6 \cdot 5! \\ &=& 6! \\ \end{array} $
proof: Induction on $k$. We want to show $(k+1)! - 1 = \sum_{i=1}^{k} i \cdot i!$, using the inductive hypothesis we have $(k+1)! - 1 = k \cdot k! + k! - 1$ which clearly holds.
This lemma tells us that if we can express every number from $0$ to $k!-1$ in $k-1$ digits, we can also express every number from $0$ to $(k+1)!-1$ with $k$ digits.
Theorem Every natural number is uniquely expressed as a Cantor number.
proof: We will show that this is true for every number $n < (k+1)!$ for all $k$ by induction on $k$. The base case is trivial. For the inductive case we note there is a unique $c_k$ such that $n - c_k \cdot k!$ is positive and $< k!$ (by Lemma) therefore $n = c_k \cdot k! + \text{the representation given by induction}$.
We can apply this proof to find the Cantor representation of a specific number. For example for $38$ the first step is to find the smallest $(k+1)! > 38$, since $4! = 24$ and $5! = 120$ we choose $k=4$. Now $38-1\cdot 4! = 14$ gives the first digit, $14 - 2 \cdot 3! = 2$ gives the second, $2 = 2!$ gives the next and the last must be $0$, therefore 38 = [1,2,1,0]
in Cantor representation.