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

This generalizes to the representation $$n = \sum_{i\ge 0} a_i b_i $$ where $b_0 = 1$ and $b_{n+1} = b_n c_n$ with $c_n > 1$ and $0 \le a_i < c_i$ - i.e., representing $n$ with digits $a_i$ to a "base" with varying ratios of consecutive digit values (phrased awkwardly, I know, but it is late and I am tired). For a usual base $B$ system, $c_i = B$ for all $i$, so $b_i = B^i$. For this system, $c_i = i$ (or maybe $c_i = i+1$, modulo my tiredness), so $b_i = i!$.
I proved too many years ago that this representation is unique iff the $c_i$ were all integers. I am sure that this was proved many years ago (probably by Euler, if not Fermat) and is in Dickson somewhere.
