For $k,m\in\Bbb Z$ show that $m! \mid (k+1)...(k+m)$

Question

how to prove that $m!$ divides product of $m$ consecutive number .

in other hand :

if we have :$k+1,...,k+m$ then $m! | (k+1)...(k+m)$ $(k,m \in \Bbb Z)$

http://math.stackexchange.com/questions/12067/the-product-of-n-consecutive-integers-is-divisible-by-n-without-using-the-prop — lab bhattacharjee, Mar 12 '13 at 18:21

score 4 · Answer 1 · answered Mar 12 '13 at 18:02

4

Hint: The number $\displaystyle \frac{(k+1)\cdots(k+m)}{m!}$ counts something :)

answered Mar 12 '13 at 18:02

Alex Youcis

score 3 · Accepted Answer · answered Mar 12 '13 at 18:50

3

Hint:

Consider the binomial coefficient $\binom{k+m}{m}$.

Good luck!

answered Mar 12 '13 at 18:50

dtldarek

2 Answers2