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 $i>m$

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 $m\lt i$.
