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?

One way to see this would be to take $f(x)=\sum_{i=0}^n {n \choose i} (-1)^i \cdot e^{xi}$ so that the $k^{th}$ derivative of $f(x)$ is $f^{(n)}(x)=\sum_{i=0}^n {n \choose i} (-1)^i \cdot i^k e^{xi}$, and $f^{(n)}(0)=\sum_{i=0}^n {n \choose i} (-1)^i \cdot i^k$. But $f(x)$ can also be written as $(1-e^x)^n$, and all the derivatives of this before the $n^{th}$ will have a factor of $(1-e^x)$, so evaluate to $0$ at $x=0$. The $n^{th}$ derivative will be $n!(1-e^x)^0\cdot(-e^x)^n$ $+$ terms with factors of $(1-e^x)$, so the sum with $k=n$ is $(-1)^n n!$.
