Factorial decomposition of integers? This question might seem strange, but I had the feeling it's possible to decompose in a unique way a number as follows:

if $x < n!$, then there is a unique way to write x as: 
       $$x = a_1\cdot 1! + a_2\cdot 2! + a_3\cdot3! + ... + a_{n-1}\cdot(n-1)!$$
       where $a_i \leq i$

I looked at factorial decomposition on google but I cannot find any name for such a decomposition.
example:
If I chose :
(a1,a2) = 


*

*1,0   -> 1

*0,1   -> 2

*1,1   -> 3

*0,2   -> 4

*1,2   -> 5


I get all number from $1$ to $3!-1$
ideas for a proof:
The number of elements between $1$ and $N!-1$ is equal to $N!-1$ and I have the feeling they are all different, so this decomposition should be right. But I didn't prove it properly.
Are there proofs of this decomposition? Does this decomposition as a name? And above all is this true ?
Thanks in advance

Your conjecture is correct. There is a straightforward proof by induction that such a decomposition always exists. Suppose that every positive integer less than $n!$ can be written in the form $\sum_{k=1}^{n-1} a_k k!$, where $0 \le a_k \le k$, and let $m$ be a positive integer such that $n! \le m < (n+1)!$. There are unique integers $a_n$ and $r$ such that $m = a_nn! + r$ and $0 \le r < n!$, and since $m < (n+1)! = (n+1)n!$, it’s clear that $a_n \le n$. Since $r < n!$, the induction hypothesis ensures that there are non-negative integers $a_1,\dots,a_{n-1}$ such that $r = \sum_{k=1}^{n-1} a_k k!$, and hence $m = \sum_{k=1}^n a_k k!$.
We’ve now seen that each of the $(n+1)!$ non-negative integers in $\{0,1,\dots,n\}$ has a representation of the form $\sum_{k=1}^n a_k k!$ with $0 \le a_k \le k$ for each $k$. However, there are only $\prod_{k=1}^n (k+1) = (n+1)!$ distinct representations of that form, so each must represent a different integer, and each integer’s representation is therefore unique.
