Constructing $\mathbb N$ from the set of factorials

Let S be the set $\{0!, 1!, 2!, \ldots\}$. Is it possible to construct any positive integer using only addition, subtraction and multiplication, and using any element in S at most once? For example:

$$ 3 = 2! + 1!$$ $$ 4 = 3! - 2! = 2! + 1! + 0!$$ $$ 146 = 4!\cdot3! + 2!$$

etc. My gut instinct says that this isn't true, but I can't see why. Something like 8076 doesn't have an obvious solution, but maybe you can get it by subtraction a huge factorial from the product of two smaller factorials or something. Or maybe there's a way of finding sets of factorials that add/subtract/multiply to 1, in which case any number can be constructed this way. I've tried finding something but haven't had much luck.

EDIT: Oops, positive integer, not positive number.

Solution 1:

Let me assume you're only allowed to use $0! = 1!$ once. In that case, all factorials past $4!$ are divisible by $24$, so working $\bmod 24$ the only numbers you're allowed to use are $1, 2, 6$, each at most once, and I am reasonably certain you cannot get any numbers congruent to $10 \bmod 24$ this way.

Edit: If you want to use both $0!$ and $1$, then all factorials past $5!$ are divisible by $120$, so working $\bmod 120$ the only numbers you're allowed to use are $1, 1, 2, 6, 24$. This time I am reasonably certain you cannot get any numbers congruent to $57 \bmod 120$. Just kidding! Every conjugacy class $\bmod 120$ is reachable.

Okay, working $\bmod 720$ the only numbers you're allowed to use are $1, 1, 2, 6, 24, 120$...

Solution 2:

Relaxing your restriction using each factorial at most once, you may find the following helpful. For every positive integer $n$ there is a positive integer $k$ and $k$ positive integers $\{c_1, \dots, c_k \}$ such that $n$ has a unique representation in the factorial basis, \begin{align} n = \sum_{l = 1}^{k} c_{l} \ l!. \end{align}