I'm trying to use the binomial coefficient:
$$\binom{x}k=\begin{cases} \frac{x^{\underline k}}{k!},&\text{if }k\ge 0\\\\ 0,&\text{if }k<0\;, \end{cases}$$
To check that ${-1\choose 0}=1$. But it doesn't make sense. I'm using, specifically:
$$\binom a k = \frac{\alpha(a-1)(a-2)\cdots(a-k+1)}{k(k-1)(k-2)\cdots 1}$$
Then I guess it would be:
$${-1 \choose 0}=\frac{(-1)(-1-0)}{0(?)}$$
The denominator should be a product of decreasing numbers from $k$ to $1$, but it's impossible to decrease from $0$ to $1$. At this point, I'm not sure if something can be made with this definition to make it work or if I should assume that $0!=1.$
$$\binom{x}{k} = \frac {\prod ~\{x - 0,~ x - 1,~ x - 2,~ \dots,~ x - (k-1)\}} {\prod~ \{k - 0,~ k - 1,~ k - 2,~ \dots,~ k - (k-1)\}}$$
When $k$ is $0$, this becomes the quotient of the products of empty sets:
$$\binom{x}{0} = \frac {\prod~ \{\}} {\prod~ \{\}}$$
And the product of the elements of an empty set is better defined as $1$, so
$$\binom{x}{0} = \frac {\prod~ \{\}} {\prod~ \{\}} = \frac 11 = 1$$