Binomial Expansion where N is negative

Comparing the formula for regular binomial expansion (n>1):

$(a+b)^n=a^n + \binom{n}1a^{n-1}b + \binom{n}2a^{n-2}b^2 +...$

to binomial expansion for negative indices, (n<1):

$(1+x)^n= 1 + nx + \dfrac{n(n-1)x^2}{2!} +...$

Does this imply that $\binom{-1}2= \dfrac{-1!}{2!(-1-2)!}=\dfrac{-1.-2}{2!}$?

Confused by the negative factorial...

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(Backstory: haven't done maths in a very long time)

For some reason that I cannot really understand (except if it is purely aesthetic) many people want to define binomial coefficients by $\binom nk=\frac{n!}{k!(n-k)!}$. But that formula is neither very efficient, nor very general (it is only valid for integers $0\leq k\leq n$). Of course binomial coefficients should be defined combinatorially, but if one looks at the basic argument that leads to the given "closed form", then it actually directly gives $$\binom nk = \frac{n(n-1)\ldots(n-k+1)}{k!}$$ (namely: choose a $k$-element subset by choosing the first element in $n$ ways, the second in $n-1$ ways, ... the last in $n-k+1$ ways, and divide that product by $k!$ to compensate for the different orderings for which each subset is obtained). Throwing in $(n-k)!$ in numerator and denominator gives the "three factorials" form.
The displayed formula is a much better one when it comes to generalising; as long as $k\in\Bbb N$, one can allow $n$ to be anything, including a negative integer, fraction, or even a formal indeterminate (in which case one obtains a polynomial of degree$~k$ with rational coefficients, taking integer values at all integers). It is sometimes useful to extend the definition so obtained by stipulating $\binom nk=0$ whenever $k$ is a negative integer, regardless of the value of$~n$ (it is more generally often useful to take $\frac1{k!}=0$ for negative integer$~k$). One reason that the generalisation is useful is the binomial formula $$(1+X)^\alpha = \sum_{k\in\Bbb N}\binom\alpha kX^k$$ that is valid as an identity of formal power series for arbitrary values of$~\alpha$, including negative integers and fractions. (Substituting $z$ for $X$ gives a converging series as right hand side whenever $|z|<1$.)
To link this with your question, it means $\binom{-1}2=\frac{-1\times-2}{2!}=1$ is correct, just skip the meaningless intermediate formula.