Factorial decomposition of integers?

You're looking for the factorial number system, also known as "factoradic". Searching should give you more results.

Yes, it's true that such a decomposition is always possible. One way to prove it is as follows: given $x < n!$, consider the $x$th permutation of some ordered set of $n$ symbols. This is some permutation $(s_1, s_2, \dots, s_n)$. Now for $s_1$ you had $n$ choices (label them $0$ to $n-1$) and you picked one, so let $a_{n-1}$ be the choice you made. For $s_2$ you had $n-1$ choices (label them $0$ to $n-2$) and you picked one, so let $a_{n-2}$ be the number of the choice you made. Etc. $a_0$ is always $0$ because you have only one choice for the last element. (This is also known as the Lehmer code.)

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.

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.