I'm trying to prove the following: $$\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}k^{n-1}=0,\ \forall n>2$$
using generating functions (or a simillar algebric way). I (barely) managed to prove this identity using “Combinatorial story” and the principle of inclusion-exclusion, and now I'm looking for a more “safe” way to prove it and improve my understanding of the different tricks and techniques used in generating functions to solve these kind of problems (this topic is new to me so I'll be really glad if you could share the thinking process in your solution too) .
my solution was to ask in how many ways we can put $n-1$ different balls in $n$ cells so that there's at least one ball on each cell.Then we can define $A_i$ as the number of ways to put the balls in different cells where the cell $i$ will be empty. then we receive that $$\left|\cap_{j=1}^{k}A_{i_{j}}\right|=\left(n-k\right)^{n-1}$$ where $\left|\cap_{j=1}^{k}A_{i_{j}}\right|$ is the set of the different partitions where the cells $i_1,...,i_k$ are empty, the we use the principle of inclusion-exclusion to complete the proof.
This isn't generating functions, but here it is anyway.
$\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}f(k) $ is the $n$-th difference of $f(k)$ and so is zero for a polynomial of degree less than $n$.
In particular, it is zero for the $n-1$ degree polynomial $f(k)=k^{n-1}$.