On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$? I've had to evaluate ${-1 \choose0}$ and then I discovered the following:
$${-1 \choose0}=\frac{(-1)!}{(-1)!0!}=\frac{(-1)!}{(-1)!}=1$$
Can I assume that $\frac{(-1)!}{(-1)!}=1$? 
 A: From the definition of the binomial coefficients we can generalize to negative numbers by applying the following method,
\begin{align}\binom{-n}{k} &= \frac{-n\cdot-(n+1)\dots-(n+k-2)\cdot-(n+k-1)}{k!}\\
&=(-1)^k\;\frac{n\cdot(n+1)\cdot(n+2)\cdots (n + k - 1)}{k!}\\
&=(-1)^k\binom{n + k - 1}{k}\\
&=(-1)^k\left(\!\!\binom{n}{k}\!\!\right)\;.
\end{align}
And by putting $n=1$ and $k=0$, the answer to your query is yes
A: Since $n!$ is infinite or undefined for negative integers, it is not really proper to use factorial on negative integers. However, there are a couple of ways to attempt to work around this problem.
We can consider $n!=\Gamma(n+1)$ and consider points very close to integers. This approach needs to be used with care. For example:
$$
\begin{align}
\binom{-1}{0}
&=\lim_{h\to0}\frac{\Gamma(h)}{\Gamma(h)\Gamma(1)}\\
&=\lim_{h\to0}\frac{\frac1h\Gamma(1+h)}{\frac1h\Gamma(1+h)\Gamma(1)}\\
&=\lim_{h\to0}\frac{\Gamma(1+h)}{\Gamma(1+h)\Gamma(1)}\\
&=\frac{\Gamma(1)}{\Gamma(1)\Gamma(1)}\\[6pt]
&=1\tag{1}
\end{align}
$$
However, if we change the approach to the limit point, things fall apart:
$$
\begin{align}
\binom{-1}{0}
&=\lim_{h\to0}\frac{\Gamma(2h)}{\Gamma(h)\Gamma(1+h)}\\
&=\lim_{h\to0}\frac{\frac1{2h}\Gamma(1+2h)}{\frac1h\Gamma(1+h)\Gamma(1+h)}\\
&=\lim_{h\to0}\frac12\frac{\Gamma(1+2h)}{\Gamma(1+h)\Gamma(1)}\\
&=\frac12\frac{\Gamma(1)}{\Gamma(1)\Gamma(1)}\\
&=\frac12\tag{2}
\end{align}
$$
This is because
$$
f(x,y)=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}\tag{3}
$$
is not continuous at $(x,y)\in\mathbb{Z}^2$ when $x\lt0$. However, if we fix $m\in\mathbb{Z}$, then we get
$$
\begin{align}
\lim_{x\to-n}f(x,m)
&=\lim_{x\to-n}\frac{\Gamma(x+1)}{\Gamma(m+1)\Gamma(x-m+1)}\\
&=\lim_{x\to-n}\frac{\frac1{x+1}\frac1{x+2}\cdots\frac1{x+n}\Gamma(x+n+1)}{\Gamma(m+1)\frac1{x-m+1}\frac1{x-m+2}\cdots\frac1{x+n}\Gamma(x+n+1)}\\
&=\lim_{x\to-n}\frac{(x-m+1)(x-m+2)\cdots x}{\Gamma(m+1)}\\
&=\frac{(-n)(-n-1)\cdots(-n-m+1)}{m!}\\
&=(-1)^m\binom{n+m-1}{m}\tag{4}
\end{align}
$$
which is the proper value given by the binomial theorem. Thus, we can consider the binomial coefficients as ratios of Gamma functions, but only if we approach along the proper path.
To address your question a bit more directly, $\frac{(-1)!}{(-1)!}=\frac{\Gamma(0)}{\Gamma(0)}$ should be equal to both $\lim\limits_{h\to0}\frac{\Gamma(2h)}{\Gamma(h)}$ and $\lim\limits_{h\to0}\frac{\Gamma(h)}{\Gamma(h)}$, but these limits are different. Therefore, not only because we are dealing with $\frac\infty\infty$, but also because these limits do not exist in general, it might be best to avoid $\frac{(-1)!}{(-1)!}$.
