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?

This is a variation using  the  coefficient   of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling $e^z=\sum_{j=0}^\infty \frac{z^j}{j!}$ we can write for instance 
\begin{align*}
n![z^n]e^{kz}=k^n\tag{1}
\end{align*}

We obtain for $n>2$
\begin{align*}
\color{blue}{\sum_{k=1}^n}&\color{blue}{(-1)^kk^{n-2}\binom{n}{k}}\\
&=\sum_{k=1}^n(-1)^k\binom{n}{k}(n-2)![z^{n-2}]e^{kz}\tag{2}\\
&=(n-2)![z^{n-2}]\sum_{k=1}^n\binom{n}{k}\left(-e^{z}\right)^k\\
&=(n-2)![z^{n-2}]\left(\left(1-e^z\right)^n-1\right)\tag{3}\\
&=(n-2)![z^{n-2}]\left((-1)^n\left(z+\frac{z^2}{2}+\cdots\right)^n-1\right)\tag{4}\\
&\,\,\color{blue}{=0}
\end{align*}

Comment:


*

*In (2) we apply the coefficient of operator according to (1)

*In (3) we apply the binomial theorem.

*In (4) we observe the coefficient of $z^{n-2}$ is zero since the left term starts with powers in $z$ greater or equal to $n$ and the constant $-1$ does not contribute since $n>2$.
Hint: Instructive examples using this and related techniques can be found in H.S. Wilf's book generatingfunctionology. See the examples following (1.2.6).
