How to show that $i^m-i(i-1)^m+\frac{i(i-1)}{1.2} (i-2)^m-...(-1)^{i-1}.i.1^m=0$? 
How to show the following? $$i^m-i(i-1)^m+\frac{i(i-1)}{1.2}
 (i-2)^m-...(-1)^{i-1}.i.1^m=0$$ (if $i>m$)

This seems really complicated.Can't spot any pattern as such :\ .Someone help me out!
P.S: I don't think the question means $i$ is iota here because it says $i>m$
 A: Theorem: For $n\gt m\ge0$, we have
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}k^m=0
$$
Proof: Suppose this holds for $n-1$, then
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}k^m
&=\sum_{k=0}^n(-1)^k\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]k^m\\
&=\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}k^m-\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}(k+1)^m\\
&=\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\left[k^m-(k+1)^m\right]\\
&=-\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\left[\sum_{j=0}^{m-1}\binom{m}{j}k^j\right]\\
&=-\sum_{j=0}^{m-1}\binom{m}{j}\color{#C00000}{\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}k^j}\\
\end{align}
$$
and each term in red is $0$ by the inductive hypothesis since $j\le m-1\lt n-1$. Therefore, the theorem holds for $n$.
All we need to show is that the theorem holds for $n=1$ and $m=0$, which is $1-1=0$.
QED

$$
\begin{align}
\sum_{k=0}^i(-1)^k\binom{i}{k}(i-k)^m
&=\sum_{k=0}^i(-1)^k\binom{i}{k}\sum_{j=0}^m(-1)^j\binom{m}{j}i^{m-j}k^j\\
&=\sum_{j=0}^m(-1)^j\binom{m}{j}i^{m-j}\color{#C00000}{\sum_{k=0}^i(-1)^k\binom{i}{k}k^j}
\end{align}
$$
where each term in red is $0$ by the Theorem since $j\le m\lt i$.
A: I suppôse that $m\geq 1$. 
Your sum seems to be
$$S=\sum_{k=0}^{i} { i \choose k}(i-k)^m (-1)^k$$
Putting $i-k=j$, this becomes
$$ S=(-1)^i \sum_{j=0}^{i} { i \choose j}(j)^m (-1)^j=(-1)^i T$$
We have 
$$\sum_{j=0}^i {i \choose j}(-1)^j x^j=(1-x)^i=P_i(x)$$
Let $\tau =x\frac{d}{dx}$. It is easy to see by induction that for $i>h$, we have  $\tau^h(P_i)(x)=Q_h(x)(1-x)^{i-h}$ where $Q_h(x)$ is a polynomial. In particular, as $i>m$, we get that $\tau^m(P_i)(1)=0$. But
$$\tau^m( \sum_{j=0}^i {i \choose j}(-1)^j x^j)=\sum_{j=0}^i {i \choose j}(-1)^j j^m x^j$$
and hence $T=0$ and we are done. 
A: Taken from this answer:
In this answer there are three proofs of
$$
\begin{align}
\sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k}
&=\left\{\begin{array}{}
1&\text{if }n=k\\
0&\text{otherwise}
\end{array}\right.\\
&=[n=k]
\end{align}
$$
where $[\dots]$ are Iverson Brackets. Furthermore, $\newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}}$
$$
\sum_{k=0}^m\binom{n}{k}\,\stirtwo{m}{k}k!=n^m
$$
where $\stirtwo{m}{k}$ are Stirling Numbers of the Second Kind. Therefore,
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}(x-k)^m
&=\sum_{k=0}^n\sum_{j=0}^m(-1)^{k-j}\binom{n}{k}\binom{m}{j}x^{m-j}k^j\\
&=\sum_{k=0}^n\sum_{j=0}^m\sum_{i=0}^j(-1)^{k-j}\binom{n}{k}\binom{m}{j}x^{m-j}\binom{k}{i}\stirtwo{j}{i}i!\\
&=\sum_{j=0}^m\sum_{i=0}^j(-1)^{n-j}\binom{m}{j}x^{m-j}\,[n=i]\,\stirtwo{j}{i}i!\\
&=\sum_{j=0}^m(-1)^{n-j}\binom{m}{j}x^{m-j}\stirtwo{j}{n}n!
\end{align}
$$
If $m\lt n$, then either $\binom{m}{j}=0$ or $\stirtwo{j}{n}=0$. If $m=n$, the only non-zero term is $j=m$.

Setting $n=x=i$ gives
$$
\begin{align}
\sum_{k=0}^i(-1)^k\binom{i}{k}(i-k)^m=0
\end{align}
$$
since $m\lt i$.
A: We give a combinatorial proof for the claim. It is contained in the following
more general
$\mathbf{Theorem.}$ The number $s_{n,m}$ of surjective maps
$f:[m]\rightarrow [n]$  is given by
$$s_{m,n}= n^m-{n\choose 1}(n-1)^m+{n\choose 2}(n-2)^m- \ldots +(-1)^n 
                  {n\choose n-1} 1^m. $$
In particular
$$ n!=
\sum_{i=0}^n 
   (-1)^i {n\choose i} (n-i)^n; $$
and if $n>m$ then $s_{n,m}=0$ (which is a proof of the original claim).
For an integer  $k\geq 1$ let  $[k]:=\{1,\ldots,k\}.$
Let $\Omega$ be the family of all maps
$f:[m]\rightarrow [n],$   and let  $A_j=\{f\in \Omega: f([m]) \subseteq [n]-\{j\}\},$  for  $j=1,\ldots, n.$  A map  $f\in \Omega$ is non-surjective  if  $f\in A_j$ for some  $j.$  For  $J\subseteq [n]$ we note
$A_J:=\bigcap_{i\in J} A_j=\{f\in \Omega: f([m]) \subseteq [n]-J\}.$
We get :  $f$ is surjective iff  $f \in \Omega-\bigcup_{j=1}^n A_j.$
Thus the number of surjective maps is by inclusion-exclusion
\begin{eqnarray*}
 |\Omega -\bigcup_{j=1}^n A_j|&=&
  |\Omega|-|\bigcup_{j=1}^n A_j|\\
  &=& n^m -  \sum_{i=1}^n (-1)^i 
 \sum_{\scriptsize \begin{array}{c} 
   I\subseteq [n] \\|I|=i
   \end{array}}  |A_I| \\
 & = & n^m- \sum_{i=1}^n (-1)^i 
\sum_{\scriptsize \begin{array}{c} 
   I\subseteq [n] \\
   |I|=i
   \end{array}} (n-|I|)^m = \sum_{i=0}^n 
   (-1)^i {n\choose i} (n-i)^m. 
\end{eqnarray*}
We thus get the first part. The second part is consequence of that the surjective functions from
$[n]$ to  $[n]$ are exactly the bijective ones; and of these there exist precisely $n!.$  Finally if $n>m$ there do not exist any surjective maps,
so then $s_{n,m}=0.$ $\Box$
