Alternating sum of binomial coefficients multiplied by index to an n-2 extent Doing research in probability modelling I obtained an identity, which is correct for $n > 2$.
$$\sum\limits^n_{i=1}(-1)^{n+i}{{n}\choose{i}}i^{n-2}=0$$
How can it be proven directly? In what literature can I find it?

My favorite proof of the probably most general form of the identity starts with the claim:
$$\frac{d^m}{dx^m}(e^x-1)^n=\sum_{k=0}^m {m \brace k}\frac{n!}{(n-k)!}
(e^x-1)^{n-k}e^{kx},\tag1
$$
which can be easily proved by induction over $m$ using the well-known summation property of the Stirling numbers of the second kind:
$$
{m \brace k-1}+k{m \brace k}={m+1 \brace k}.
$$
Now consider the obvious equality:
$$
(e^x-1)^n=\sum_{k=0}^n (-1)^{n-k}\binom nk e^{kx}.\tag2
$$
Differentiating both sides $m$ times with respect to $x$ and substituting $x=0$ in the resulting expression:
$$\sum_{k=0}^m {m \brace k}\frac{n!}{(n-k)!}
(e^x-1)^{n-k}e^{kx}=\sum_{k=0}^n (-1)^{n-k}\binom nk k^me^{kx},\tag3
$$
one obtains:
$$
\boxed{{m \brace n}n!=\sum_{k=0}^n (-1)^{n-k}\binom nk k^m}\tag4
$$
since only $k=n$ can provide a non-zero term in the left hand side sum upon the substitution. Particularly for $m<n$ the sum is $0$ and your claim follows.
