Prove that $ \sum_{i=0}^n {n \choose i} \cdot (-1)^i \cdot i^n =(-1)^n n!$ and $ \sum_{i=0}^n {n \choose i} \cdot (-1)^i \cdot i^k = 0$ for $kFor natural any number $n$, prove that $\sum_{i=0}^n{n \choose i}\cdot (-1)^i \cdot i^n =(-1)^n n!$ and for whole number $k<n,\,\sum_{i=0}^n {n \choose i} \cdot (-1)^i \cdot i^k =0$.
I know that $\sum_{i=0}^n{n \choose i}\cdot (-1)^i \cdot i^k = 0.$
For $k=0,1,2$ as it is expansion of $(x+1)^n,\,n(x+1)^{n-1},\,n(n+1)(x+1)^{n-2}$ at $x=-1$. But I am not able to prove for $k>2$. Can someone help me out?

A variation: We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. We also recall the series expansion $e^z=1+z+\frac{1}{2}z^2+\cdots=\sum_{n=0}^{\infty}\frac{z^n}{n!}$.

We obtain for non-negative integer $k$ with $0\leq k\leq n$:
\begin{align*}
\color{blue}{\sum_{i=0}^n\binom{n}{q}(-1)^qq^k}
&=\sum_{q=0}^n\binom{n}{q}(-1)^qk![z^k]e^{qz}\\
&=k![z^k]\sum_{q=0}^n\binom{n}{q}\left(-e^z\right)^q\\
&=k![z^k]\left(1-e^z\right)^n\\
&=k![z^k]\left(-z-\frac{z^2}{2}-\cdots\right)^n\\
&\,\,\color{blue}{=\begin{cases}
0&\qquad0\leq k< n\\
(-1)^nn!&\qquad k=n
\end{cases}}
\end{align*}
and the claim follows.

