If we know that
$$\lim_{m \to \infty} (-m+\frac 1 2) =0$$
when $m \in \Bbb z$
Can we say that:$$(-\infty)!=0$$
Actually by looking to the graph of gamma function the curve tends to infinity for the negative integers.
But what happen if $m \rightarrow \infty$ and $m \in \Bbb Z$ for:
$$\lim_{m \to -\infty} (-m)!$$
I was thinking it will be the same , but the following steps gave me zero value for it, as below.
First we have fact $$\lim_{m \to \infty} (\frac 1 m)!=1$$
The limit can be written as $$\lim_{m \to \infty} (\frac 1 m-1)!\frac 1 m =1$$
by repeating the process we get $$\lim_{m \to \infty} ](\frac 1 m-m-1)! \left [ \frac 1 m \, (\frac 1 m-1)(\frac 1 m-2)...(\frac 1 m-m) \right ]=1$$
Now we have
$$ \lim_{m \to \infty}(\frac 1 m-1)(\frac 1 m-2)...(\frac 1 m-m)= \lim_{m \to \infty}(-1)^m(\frac 1 m+m)!$$
by substitute it in the equation.
$$\lim_{m \to \infty} (\frac 1 m-m-1)! \left [\frac 1 m \,(-1)^m(\frac 1 m+m)!\right ] =1$$
$$\lim_{m\to \infty}(\frac 1 m-m-1)!=\lim_{m \to \infty} \frac m {(-1)^m(\frac 1 m+m)!}$$
$$\lim_{m\to \infty}(\frac 1 m-m-1)!=\lim_{m \to \infty} \frac {(\frac 1 m +m)-\frac 1 m} {(-1)^m(\frac 1 m+m)!}$$
$$\lim_{m\to \infty}(\frac 1 m-m-1)!=\lim_{m \to \infty} \left [\frac {1} {(-1)^m(\frac 1 m+m-1)!}-\frac {1} {(-1)^m(\frac 1 m+m)!\,m}\right ]$$
$$(-\infty)!=0$$
So is that true?... actually I don't know.
The question is simple.
The statement
$$n! = n\cdot (n-1)!$$
only holds when $n\in\mathbb{N}$
In your case you have not a natural number a priori (you considered indeed integers, and factorial is not defined for negative naturals), since for $m\in\mathbb{Z}$, the quantity $\frac{1}{m} \notin \mathbb{Z}$ so you cannot write
$$\left(\frac{1}{m}\right)! = \frac{1}{m}\left(\frac{1}{m} -1 \right)!$$
$\Gamma[-\infty]$
By the definition:
$$\Gamma[x] = \int_0^{+\infty} t^{x-1}e^{-t}\ \text{d}t$$
But what we have is
$$\Gamma[-\infty] = \text{Indeterminate}$$