How was this simplification done?

Question

\begin{align*} \begin{pmatrix} -2\\ n\end{pmatrix}&=\frac{(-2)(-3)(-4)\dots(-2-n+1)}{n!}\\[5pt] &= (-1)^n\frac{2\times3\times4\times\dots n(n+1)}{n!}\\[5pt] &=(-1)^n(n+1)\\[5pt] \end{align*}

While reviewing the binomial theorem, I came across this example problem and I feel like this simplification was not clearly explained. It seems like -1 was factored out but what happens to the expression it is multiplied by. How is it written in such a way? Also, why is $-1$ raised to the power of $n$...$(-1)^n$. I do not find this intuitive and felt like a step was skipped although that might not be the case.

Overall, I just want to understand what the process is here.

Answer 1

I like using the product notation so everything is explicit.

$\begin{array}\\ \binom{-2}{n} &=\dfrac{\prod_{k=0}^{n-1} (-2-k)}{n!}\\ &=\dfrac{(-1)^n\prod_{k=0}^{n-1} (2+k)}{n!}\\ &=\dfrac{(-1)^n\prod_{k=2}^{n+1} k}{n!}\\ &=\dfrac{(-1)^n\prod_{k=1}^{n+1} k}{n!} \qquad\text{(adds the term with } k=1)\\ &=\dfrac{(-1)^n(n+1)!}{n!}\\ &=(-1)^n(n+1)\\ \end{array} $

More generally, if $m$ is a positive integer,

$\begin{array}\\ \binom{-m}{n} &=\dfrac{\prod_{k=0}^{n-1} (-m-k)}{n!}\\ &=\dfrac{(-1)^n\prod_{k=0}^{n-1} (m+k)}{n!}\\ &=\dfrac{(-1)^n\prod_{k=m}^{n+m-1} k}{n!}\\ &=\dfrac{(-1)^n\prod_{k=1}^{n+m-1} k}{n!\prod_{k=1}^{m-1}k} \qquad\text{(adds the term with } k=1 \text{ to } m-1)\\ &=\dfrac{(-1)^n(n+m-1)!}{n!(m-1)!}\\ &=(-1)^n\binom{n+m-1}{m-1}\\ \end{array} $

Answer 2

Suppose $m>n$. We can write simplify $$\dbinom mn=\dfrac{m!}{n!\,(m-n)!}=\frac{m(m-1)(m-2)\dots(m-n+1)}{n!}.$$

By analogy, if $m<n$, we can set $$\dbinom mn=\frac{m(m-1)(m-2)\dots(m-n+1)}{n!}.$$ In particular, if $m<0$, the numerator has all its factors negative, so we can write it as \begin{align}&-|m|\cdot -\bigl(|m|+1\bigr)\cdot -\bigl(|m|+2\bigr)\dots -\bigl(|m|+n-1\bigr)=\\[1ex] &(-1)^n|m|\bigl(|m|+1\bigr)\bigl(|m|+2\bigr)\dots\bigl(|m|+n-1\bigr) \end{align} since there are $n$ factors.