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 $i>m$

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 $s_{n,m}=0.$ $\Box$
