Basically, I need to show the number of functions from a set $A$ with $m$ elements onto the set $B$ with $n$ elements is $n!$ whenever $m=n$. It is easy to argue that each onto function is a bijection (permutation) under the scenario $m=n$ and thereby the desired number is $n!$. But I need to show using the formula $$n^m -n(n-1)^m+~^nC_2(n-2)^m- \cdots+(-1)^{n-2}~^nC_{n-2}2^m+(-1)^{n-1}~^nC_{n-1}=\sum_{i=0}^{n-1}(-1)^i~{}^nC_i(n-i)^m,$$ which gives the number of onto functions from $A$ to $B$ whenever $m\geq n$.
I tried to simplify it and reached at $$n! \sum_{i=0}^{n-1}(-1)^i \frac{(n-i)^m}{i!(n-i)!}.$$ How to make $$\sum_{i=0}^{n-1}(-1)^i \frac{(n-i)^m}{i!(n-i)!}=1,$$ for $m=n$?