For 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?
The binomial formula gives $$ (x+1)^n = \sum_{i=0}^n {n \choose i} x^i .$$ Apply the differential operator $x \frac{d}{dx}$ to both sides $1 \leq k \leq n$ times to get $$ \frac{n!}{(n-k)!} x^k (x+1)^{n-k} + (x+1)p_k(x) = \sum_{i=0}^n {n \choose i} i^k x^i , $$ where $p_k (x)$ is some polynomial. Plugging in $x=-1$ gives the identity.
Actually, for $k =n$ the resulting identity differs from yours by a sign: $(-1)^n n! = \sum_{i=0}^n {n \choose i} i^n (-1)^i$.
You could also do this combinatorially. Paint $k$ walls in $n$ colours. Then the number of ways of doing this using all $n$ colours is, by the inclusion-exclusion principle, $n^k$ [any wall in any colour] $-{n \choose 1} \cdot (n-1)^k$ [pick one colour and count how many ways you can paint the walls omitting this colour; you must exclude these] $+{n \choose 2} \cdot (n-2)^k$ [add back the ways you excluded twice because they omit two colours] $+ ...$. Hence the number of ways is given by $\sum_{i=1}^n{n \choose i} (-1)^{n-i}i^k$, which is $(-1)^n$ times your sum. However, there are no ways of painting fewer than $n$ walls in $n$ colours using all the colours, so all the sums are zero, except where $k=n$, when there are $n!$ ways of doing it.
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 $(-1)^n n!$.
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.
We can use combinatorial proof here.
Method 1
Let $i$ students be chosen out of a class of $n$ to be graduation committee. Subsequently, each of $k$ tasks need to be assigned to exactly one committee member. Each committee member can handle multiple tasks. The difference between number of possible arrangement where the committee consist of even number of people vs odd number of people is given by the original equation.
$$ \sum_{i=0}^{n}\binom{n}{i}\cdot(-1)^{i}\cdot i^{k} $$
Method 2
Each of $k$ tasks is assigned to exactly one students and the assigned students automatically become a committee member. If there are less tasks than students ($k<n$), there will always be remaining students that can still be committee member. Since the number of this additional committee member is equally likely to be odd or even, the difference mentioned in method 1 must be zero.
It there are equal number of tasks and students ($k=n$), it is possible that each student handle exactly one task ($n!$ possibilities). Since there is no students left to be made committee, the difference is not zero but instead $(-1)^{n}n!$.
Conclusion
$$ \sum_{i=0}^{n}\binom{n}{i}\cdot(-1)^{i}\cdot i^{k} = 0, \phantom{xx} k=0,1,2,...,n-1 $$
$$ \sum_{i=0}^{n}\binom{n}{i}\cdot(-1)^{i}\cdot i^{n} = (-1)^{n}n! $$
Here is another approach for $k<n$.
Define the difference operator $\Delta$ so that $(\Delta f)(n)=f(n)-f(n-1)$. Note that $\Delta=1-\mathcal{P}$, where $(\mathcal{P}f)(n)=f(n-1)$. Then $$ \Delta^m=(1-\mathcal{P})^m=\sum_{i=0}^{m}{\binom{m}{i}(-1)^i\mathcal{P}^i}, $$ so $$ \begin{split} (\Delta^m\!f)(n)&=\sum_{i=0}^{m}{\binom{m}{i}(-1)^i f(n-i)}=\sum_{i=0}^{m}{\binom{m}{i}(-1)^{m-i}f(n-m+i)}\\ &=(-1)^m\sum_{i=0}^{m}{\binom{m}{i}(-1)^i f(n-m+i)}. \end{split} $$ Let $m=n$, then $$ (\Delta^n\!f)(n)=(-1)^n\sum_{i=0}^{m}{\binom{n}{i}(-1)^i f(i)}. $$ If $f(n)$ is a polynomial (such as $n^k$ in our example), then every application of $\Delta$ yields a polynomial of a lower degree. Thus, applying $\Delta^n$ to any polynomial of degree less than $n$ yields a polynomial of degree below zero, i.e. the constant $0$.
From this question we can get an alternative answer: In the expression $$ \sum_{i=1}^n\left(\left. x_i^k\right/ \prod_{\substack{1\leqslant j\leqslant n \\ j\neq i}}(x_i-x_j)\right) = \begin{cases}0, & \text{if } 0\leqslant k < n-1; \\ 1, & \text{if } k=n-1; \\ \sum_{i=1}^nx_i, & \text{if } k=n.\end{cases} $$ if we take $x_i=i$ and multiply the expression by $n!$ we get $$ \sum_{i=1}^n\left( \frac{n!}{\prod_{\substack{1\leqslant j\leqslant n \\ j\neq i}}(i-j)}i^k\right) = \begin{cases}0, & \text{if } 0\leqslant k < n-1; \\ n!, & \text{if } k=n-1; \\ \frac{n(n+1)}{2}n!, & \text{if } k=n.\end{cases} $$ and $$ \sum_{i=1}^n {n\choose i}(-1)^{n-i}i^{k+1} = \begin{cases}0, & \text{if } 1\leqslant k+1 < n; \\ n!, & \text{if } k+1=n; \\ \frac{n(n+1)}{2}n!, & \text{if } k+1=n+1.\end{cases} $$ and $$ \sum_{i=0}^n {n\choose i}(-1)^{i}i^{k} = \begin{cases}0, & \text{if } 1\leqslant k < n; \\ (-1)^n n!, & \text{if } k=n; \\ (-1)^n\frac{n(n+1)}{2}n!, & \text{if } k=n+1.\end{cases} $$
