Do negative binomials imply negative factorials exist? I've seen the following identity:
$$\binom{-n}{k} = (-1)^k\binom{n+k-1}{k}$$
So I tried to derive it, assuming negative factorial was a real concept, having it extend down to negative infinity:
$$\binom{-n}{k} =  \frac{(-n)!}{k!(-n-k)!}$$
Expanding to
$$\frac{(-n)(-n-1)(-n-2)...}{k!(-n-k)(-n-k-1)(-n-k-2)...}$$
Reframing to
$$\frac{(-(n))(-(n+1))(-(n+2))...}{k!(-(n+k)-(n+k+1)-(n+k+2)...)}$$
Simplifying to
$$\frac{(-(n))(-(n+1))(-(n+2))...(-(n+k-2))(-(n+k-1))}{k!}$$
Then
$$(-1)^k\frac{(n)(n+1)(n+2)...(n+k-2)(n+k-1)}{k!}$$
Which also equals
$$(-1)^k\frac{(n-1)!(n)(n+1)(n+2)...(n+k-2)(n+k-1)}{(n-1)!k!}$$
or
$$(-1)^k\frac{(n+k-1)!}{k!(n-1)!}$$
or
$$(-1)^k \binom{n+k-1}{k}$$
I am a little surprised that it seemed to work. Was this a happy coincidence? What does this imply for the factorial if I was able to treat it as something going down to negative infinity? 
 A: TLDR: Factorials of negative integers don't exist under usual definitions.

Proving the identity:
Using the definition of the generalized binomial coefficient:
$$\binom{n}{k}=\frac{n^{\underline{k}}}{k!}~~~~~~\text{where}~n\in\Bbb C, k\in\Bbb N$$
where $n^{\underline{k}}=\underbrace{n(n-1)(n-2)\cdots(n-k+2)(n-k+1)}_{k~\text{terms}}$ is the falling factorial.
We have:
$\begin{array}{rlr}\binom{-n}{k}&=\frac{(-n)^{\underline{k}}}{k!}\\
&=\frac{(-n)(-n-1)(-n-2)\cdots(-n-k+2)(-n-k+1)}{k!}\\
&=\frac{(-n)(-n-1)(-n-2)\cdots(-n-k+2)(-n-k+1)}{k!}\cdot (-1)^{k}\cdot (-1)^k&\text{as}~((-1)^k)^2=1\\
\color{red}{(\dagger)}&=\frac{(n)(n+1)(n+2)\cdots(n+k-2)(n+k-1)}{k!}\cdot (-1)^k&\text{distributing}~(-1)^k~\text{across numerator}\\
&=\frac{(n+k-1)(n+k-2)\cdots (n+2)(n+1)(n)}{k!}\cdot (-1)^k&\text{rearranging numerator}\\
&=\frac{((n+k-1))((n+k-1)-1)\cdots ((n+k-1)-k+2)((n+k-1)-k+1)}{k!}\cdot(-1)^k&\text{rewriting terms in numerator}\\
&=\frac{(n+k-1)^{\underline{k}}}{k!}\cdot (-1)^k&\text{recognizing falling factorial}\\
&=\binom{n+k-1}{k}\cdot (-1)^k&\text{generalized binomial coeffecient}\end{array}$
Technically, we could have stopped sooner at $\color{red}{(\dagger)}$ had we noted that $\frac{n^{\overline{k}}}{k!}=\binom{n+k-1}{k}$ where $n^\overline{k}$ is the rising factorial, $n^\overline{k}=n(n+1)(n+2)\cdots(n+k-1)$.
Comments about involvement of negative factorials:
Factorials of negative numbers do not appear in the above proof or definitions at all.  In fact, going to the gamma function, a generalization of factorials to all complex numbers, one has that $\Gamma(n)$ is a pole for all negative integers $n$ (similar to a division by zero error).
Ignoring this, if you try to look at something like $(-5)!$ as $(-5)(-6)(-7)(-8)\cdots=\lim\limits_{n\to\infty}\prod\limits_{k=5}^n (-k)$, as an infinite product it will clearly diverge as the related infinite series $\lim\limits_{n\to\infty}\sum\limits_{k=5}^n\ln(-k)$ diverges.

As for your algebra implying the negative factorial's existence, that would be a happy accident due to the cancellations involved.  In terms of logical implications, $P\Rightarrow \text{True}$ does not imply that $P$ is true.  By assuming it exists, you concluded a true statement, but that by itself does not prove that it exists.
