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 $i>m$

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. 
