Well, $n!$ is for integer $n < 0$ not defined — as yet.
So the question is:
How could a sensible generalization of the factorial for negative integers look like?
Clearly a good generalization should have a clear combinatorial meaning which combines well with the nonnegative case.
It's not that it's not defined... Actually it has been defined more than it should have. There are plenty of functions that interpolate the factorials, some of them extend to the negative integers as well. Hadamard's Gamma function is entire, logarithmic single inflected factorial function is another example. But on the other hand, for some mysterious reason, the nice property that we want an extension of the factorial to enjoy is log-convexity. The Bohr-Mollerup-Artin Theorem tells us that the only function which is logarithmically convex on the positive real line and satisfies $f(z)=zf(z-1)$ there (also $f(1)=1$ and $f(z)>0$) is the Gamma function. Unfortunately the gamma function doesn't extend to negative integers, and that is why I guess people don't really care that much for defining them as they know that no "good" answer can be found.
I think it's worth pointing out here that, if $a\ge0$, then, near z = -a, we have
$$ \Gamma(z) = (-1)^a {1 \over a!} {1 \over {z+a}} + O(1) $$
and so it might be tempting to say that, in some sense,
$$ \Gamma(-a) = (-1)^a {1 \over a!} \infty $$
where the symbol $\infty$ represents the rate at which $\Gamma$ blows up near the pole at $a = 0$. That is, $\Gamma(0) = \infty, \Gamma(-1) = -\infty, \Gamma(-2) = \infty/2, \Gamma(-3) = -\infty/6$, and so on.
In particular, this interpretation might work in some formula in which $\Gamma$ evaluated at nonpositive integers appears in both the numerator and the denominator, and the symbol $\infty$ can be canceled to yield a real number.
For a related paper see D. Loeb, Sets with a negative number of elements, Adv. Math. 91 (1992), 64–74.
If you're wanting to compute factorials as an intermediate step to computing binomial coefficients, you may find a more satisfactory answer to your question. See this chart for determining how to compute binomial coefficients for general arguments.
I think, a sensical definition stems from the generalization of the triangle of eulerian numbers.
For positive integer indexes the rows sum to factorials, and even if the rows are interpolated to fractional indexes (based on the closed-form-formula for the direct computation) the rowsums are fractional factorials or gamma values.
Thus I assume the extension of the eulerian triangle to negative indexes gives the answer to a sensical definition of the factorials at negative parameters.
For instance, we get $$ \small \begin{array}{r|lllll|lll} \text{index $k$} & \\ \hline\\ &&&& \text{ extension to negative indexes} \\ \cdots &\cdots \\ -4& 1 & 3+\frac 1{16} & 6+\frac 3{16}+\frac 1{81}& 10+\frac 6{16}+\frac 3{81}+\frac 1{256}&\ldots &\tiny \sum \overset ?= & -4! \\ -3& 1 & 2+\frac 18 & 3+\frac 28+\frac 1{27}& 4+\frac 38+\frac 2{27}+\frac 1{64}&\ldots &\tiny \sum \overset ?= & -3! \\ -2& 1 & 1+\frac 14& 1+\frac 14+\frac 19& 1+\frac 14+\frac 19 + \frac 1{16}& \ldots &\tiny \sum \overset ?= & -2! \\ -1& 1 & 0+\frac 12 & 0+0+\frac 13 & 0+0+0+\frac 14 & \ldots &\tiny \sum \overset ?= & -1! \\ \hline \\ &&&& \text{ triangle of Eulerian numbers} \\ 0& 1 & . & . & . &. &\tiny\sum = & 0! \\ 1& 1 & . & . & . &. &\tiny\sum = & 1! \\ 2& 1 & 1 & . & . &. &\tiny\sum = & 2! \\ 3& 1 & 4 & 1 & . &. &\tiny\sum = & 3! \\ 4& 1 & 11 & 11 & 1 &. &\tiny\sum = & 4! \\ \cdots & \cdots \\ \end{array} $$ and so on.
I have a more involved discussion in a hobby-treatise about the Eulerian-triangle here.
Hadrian Ulgenes David Peter give the following answer in Series and Product Representations of Gamma and Pseudogamma Functions, Theorem 5:
The function
\begin{equation} \Lambda(x)=\prod_{n=1}^{\infty}\prod_{k=1}^{n} k^{\frac{\left(-1\right)^{n+k}\left(2k-1\right)}{\left(n-k\right)!\left(k+n-1\right)!}\frac{\left(x+n-1\right)!}{\left(x-n\right)!\left(2n-1\right)}} \end{equation}
interpolates the factorial at the positive integers, interpolates the reciprocal factorial at the negative integers, and converges for the entire real axis.
My question was intended somewhat along the line: Assume the Gamma function is not yet invented and Goldbach asks you the question: "What is (-n)! ?" You know that Goldbach expects a combinatorial answer in the domain of integer or rational numbers. What would you answer? I will give my answer in this sense.
Looking at Graham, Knuth, and Patashnik's Concrete Mathematics, Table 253, the combined Stirling triangles in their dual form, we see: If we sum the columns in this triangle for k < 0 we get the factorial numbers, if we sum the rows for k > 0 we get the Bell numbers.
What about saying the Bell numbers are the factorial numbers at negative integers? Is the answer encoded in one of the most important triangles in combinatorics?
See what Knuth (Two notes on notation) says about the origin of this duality (table on page 11).
{120}
. {24}
1, . {6}
10, 1, . {2}
35, 6, 1, . {1}
50, 11, 3, 1, . {1}
24, 6, 2, 1, 1, .
0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 1,.....{1}
0, 0, 0, 0, 0, 0, 1, 1,....{2}
0, 0, 0, 0, 0, 0, 1, 3, 1,....{5}
0, 0, 0, 0, 0, 0, 1, 7, 6, 1,....{15}
0, 0, 0, 0, 0, 0, 1, 15, 25, 10, 1,....{52}
As 0! and 1! = 1 ,
2! = 2,
3! = 6 and so on
Can we not have
-1! = -1
-2! = 2 = -1 X -2
-3! = -6 = -1 X -2 X -3
-4! = 24 = -1 X -2 X -3 X -4
-5! = -120 = -1 X -2 X -3 X -4 X -5
...and so on?
It would produce some sort of series but would they be of any use anywhere?
According do the definition of factorial, $1 = 0! $ and $ 0! = -1! * 0$. So, first negative integer factorial is $$-1! = 1/0 = \infty$$. I am not sure why it should be a negative infinity. Possibly because zero can be very small negative number as well as positive. I cannot derive the sign. But, I can prove that other integer negatives are also infinities.
Take -2! * -1 = -1!. It follows that $-2! = -1!/-1 = -\infty$.
Next, -3! * -2 = -2! whereupon, $-3! = -2!/-2 = {-1!\over (-1)(-2)} = +\infty$.
Generally, we see that all factorials are infinities with alternating sign, $$-n! = {-1! \over (-1)(-2)\cdots(1-n)} = \infty/(-1)(-2)\cdots = (-1)^n\infty$$
