TITLE: How was this simplification done?
QUESTION [0 upvotes]: \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.

REPLY [1 votes]: 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.
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