Curious Binomial identity that came up in derivative pricing How can I show that, for a given integer $n$, that the following identity is true?
$$\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]=-(-1)^m$$
This came up in a derivation of arbitrary derivatives of the Black-Scholes function.  It is part of a larger equation for the $n^{th}$ order derivative
$$\sum_{m=0}^{n-2}\frac{H_m(d)}{(\sigma\sqrt{\tau})^m}\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]$$
which, in order to make contact with a known result (Carr, Peter. (2000). Deriving Derivatives of Derivative Securities. Journal of Computational Finance. 4. 101 - 128.
https://ieeexplore.ieee.org/document/844609) this identity must hold.  I've also checked that it's true in sagemath.
 A: $$
\begin{align}
&(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\\
&=\sum_{k=m}^{n-1}(-1)^{k+1}\binom{n}{n-k-1}\binom{k}{k-m}\tag1\\
&=(-1)^{m+1}\sum_{k=m}^{n-1}\binom{n}{n-k-1}\binom{-m-1}{k-m}\tag2\\
&=(-1)^{m+1}\binom{n-m-1}{n-m-1}\tag3\\[9pt]
&=(-1)^{m+1}[n\ge m+1]\tag4
\end{align}
$$
Explanation:
$(1)$: $(-1)^n\binom{n-1}{m}$ becomes the $m=n-1$ term of the sum
$\phantom{\text{(1):}}$ furthermore, $\binom{k}{m}=\binom{k}{k-m}$
$(2)$: $\binom{k}{k-m}=(-1)^{k-m}\binom{-m-1}{k-m}$ (negative binomial coefficients)
$(3)$: Vandermonde's Identity
$(4)$: Iverson Brackets
A: Have an alternative solution that takes a different approach than the one here. Consider mine more opaque and involved, but worth sharing for the fun of it.
We start with Eq. 1.23 in this reference changing the labeling for convenience
$$\sum_{k=0}^{n'}(-1)^k \binom {n'} k \binom {k} m x^k=(-1)^mx^m(1-x)^{n'-m} \binom{n'}{m} $$
and apply $\binom n k = \frac{n}{k} \binom {n-1} {k-1}$ as
$$\sum_{k=0}^{n'}(-1)^k \frac{k+1}{n'+1}\binom {n'+1} {k+1} \binom {k} m x^k=(-1)^mx^m(1-x)^{n'-m} \binom{n'}{m}. $$
Substitution $n=n'+1$ then yields
$$\sum_{k=0}^{n-1}(-1)^k \frac{k+1}{n}\binom {n} {k+1} \binom {k} m x^k=(-1)^mx^m(1-x)^{n-m-1} \binom{n-1}{m}. $$
Integrate both sides, assume that constants are 0, $x=1$ to get
$$\frac{1}{n}\sum_{k=0}^{n-1}(-1)^k \binom {n} {k+1} \binom {k} m=\frac{(-1)^m}{n}. $$
Integral on the RHS is tricky in general, but evaluates nicely in the present case.
Now we can split off the $n-1$ term from the sum and multiply by $n$ to obtain
$$(-1)^{n-1} \binom {n-1} m + \sum_{k=0}^{n-2}(-1)^k \binom {n} {k+1} \binom {k} m=(-1)^m. $$
Finally, we multiply by $-1$, note that binomial coefficients with $k < m$ in the sum are 0 and get
$$(-1)^{n} \binom {n-1} m - \sum_{k=m}^{n-2}(-1)^k \binom {n} {k+1} \binom {k} m=-(-1)^m. $$
