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My homework assignment was essentially to prove the following binomial coefficient identity:

Prove $$\begin{pmatrix} -n \\ k\end{pmatrix} = (-1)^k \begin{pmatrix} n+k-1 \\ k\end{pmatrix}$$

Using the definition $\begin{pmatrix} n \\ k\end{pmatrix} := \prod_{j=1}^k \frac{n+1-j}{j}$ I get

$$\prod_{j=1}^k \frac{-n+1-j}{j} = (-1)^k \prod_{j=1}^k \frac{n+k-1+1-j}{j}$$

Which is equivalent to

$$\prod_{j=1}^k -n+1-j = \prod_{j=1}^k -n-k+j$$

Now the $1-j$ in the first term and the $-k+j$ in the second term create the same set, but in the opposite direction: $1-j$ goes from $0$ to $1-k$, while $-k+j$ goes from $1-k$ to $0$, so both products are the same.

But how do I prove this in a formal way?

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    $\begingroup$ Just a $\LaTeX$ tip: \binom{n+k-1}{k} is less typing than \begin{pmatrix} ... \end{pmatrix} (and may have better spacing properties, but I don't know if they do differ in that respect). $\endgroup$ – Daniel Fischer Nov 8 '14 at 20:08
  • $\begingroup$ Strange question... by definition, if $n<k$ therefore $\binom{n}{k}=0$. $\endgroup$ – idm Nov 8 '14 at 20:19
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\begin{align} \binom{-n}{k} &= \frac{(-n - k + 1)(-n - k + 2)\cdots (-n)}{k!}\\ & = (-1)^k \frac{(n + k - 1)(n + k - 2)\cdots (n)}{k!} \\ & = (-1)^k \binom{n+k-1}{k}. \end{align}

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