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.
 A: 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}
A: 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 $1, 1, 2, 6, 24, 120$... 
A: The numbers you can construct are of the form $\sum_{j=0}^nc_jj!$ with $c_j\in\{-1,\,0,\,+1\}$. Imagine you want to construct a number smaller than $4165$. Since $7!-\sum_{j=0}^6j!=4166$, you can only use the numbers $k!$ with $0\leq k\leq 6$. That's $7$ numbers. Using your formula you can construct $3^7=2187$ numbers this way. Sadly, $2187<4165$ so the answer is no.
