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

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Theorem

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

Then:

$\paren {-1}^n \dbinom {-n} {k - 1} = \paren {-1}^k \dbinom {-k} {n - 1}$


Proof

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


Sources