# -1^n by -n choose k-1 equals -1^k by -k choose n-1

## Theorem

Let $n, k \in \Z_{\ge 0}$.

Then:

$\left({-1}\right)^n \dbinom {-n} {k - 1} = \left({-1}\right)^k \dbinom {-k} {n - 1}$

## Proof

 $\displaystyle \left({-1}\right)^n \dbinom {-n} {k - 1}$ $=$ $\displaystyle \left({-1}\right)^n \left({\left({-1}\right)^{-n - \left({k - 1}\right)} \binom {- \left({k - 1 + 1}\right)} {- \left({n - \left({k - 1}\right)}\right)} }\right)$ Moving Top Index to Bottom in Binomial Coefficient $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{- \left({k - 1}\right)} \binom {- k} {- \left({n + 1}\right) - k}$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^k \binom {- k} {- k - \left({n + 1}\right)}$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^k \binom {- k} {n - 1}$ Symmetry Rule for Binomial Coefficients

$\blacksquare$