How can I prove that $m!|(k+1)(k+2)...(k+m)$, where $m$ is a positive integer and $k$ is bigger or equal to $0$ I took a course of discrete mathematics the last semester. There were some exercises in the book that we used that where really hard, but I did the majority of them. the problem above is one that I have not been able to do. I would like to see the proof so it can help me to solve further problems.
 sorry for my english.
 A: When successively drawimg $m$ elements from a set $S$ of size $|S|=k+m$ elements (without repetitions), there are $(k+m)(k+m-1) \cdots (k+1)$ possible (ordered) results. These results can be partitioned into groups of $m!$ elements each, where two selections belong to the same group if and only if they contain the same $m$ elements. 
These groups obviously correspond to the different subsets of $S$ that contain exactly $m$ elements. The number of these subsets is known as the binomial coefficient $$\binom{m+k}{m}=\frac{(k+1)(k+2) \cdots (k+m)}{m!}$$
A: Let $[x]$ denote the largest integer not exceeding $x.$ Let $\pi$ be the set of primes.
For $m\ge p\in  \pi:$
(i). Let $p^{C(p)}$ be the largest power of $p$ that divides $\prod_{j=1}^m(k+j).$
(ii). Let $p^{D(p)}$ be the largest power of $p$ that divides $m!.$
(ii). Show that the number of multiples of $p^n$ within the range $k+1,...,k+m$ is at least  the number of multiples of $p^n$ that are $\le m,$ which is $[m/p^n].$ Conclude that $C(p)\ge\sum_{n\in \Bbb N}[m!/p^n].$ (Only a finite number of terms in that summation are non-zero.)
(iii). Show that $\sum_{n\in \Bbb N}[m!/p^n]=D(p).$ Therefore $\prod_{j=1}^m(k+j)$ is divisible by $\prod_{m\ge p\in \pi}p^{D(p)}=m!\,. $
