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

You can also reason as follows : suppose you've shown for some integer $n$ that every integer $\in\lbrace0,\dots,n!-1\rbrace$ has a unique decomposition as you suggest. Take $k\in\lbrace0,\dots,(n+1)!-1\rbrace.$ Write $$k=q\cdot n!+r$$ the euclidean division of $k$ with respect to $n!$. Necessarily you have $0\leq q < n+1$. 
This gives you an expression you want, for $0\leq r<n!$ has, by hypothesis, an expression involving only factorials up to $(n-1)!$ .
Finally, to show uniqueness, you can again use your hypothesis to deduce that if $k=a_n\cdot n!+ \sum_0^{n-1} a_i\cdot i!,$ then $0\leq \sum\dots< n!-1$ by hypothesis, and this tells you that this decomposition is the euclidean division of $k$ by $n!$. So by uniqueness of the euclidean division, it's the one decomposition we defined earlier.
