Consider the binomial coefficient $\binom{m}{n} = \frac{m(m-1)...(m-n+1)}{n!}$. It is an integer for all m, so we are done for positive sequences of integers. Clearly this proof works also for strictly negative sequences, and in the case where one of the integers is 0, the product will be 0, and hence divisible by anything.

Let the consecutive +ve integers be (r+1), (r+2)......(r+n) Then we have $ (r+1)(r+2)(r+3).......(r+n) =\frac{r!(r+1)(r+2)(r+3).......(r+n)}{r!} =\frac{(r+n)!}{r!} =\frac{(r+n)!n!}{r!n!} =\frac{(n+r)!n!}{n![(n+r)-n]} =C(n+r,n)n!$ Hence the result!

But it's so clear, since when we have $n$ consecutive integers each of them are different in $(mod k)$ where k=<n so at least one term must be $0$ in $(mod k)$

Although your way to express it is not altogether correct, it is true that for any $1\leq k \leq n$ at least one term is divisible by $k$. However, this does not guarantee the product of the terms is divisible by $n!$; think about it ... for example, looking at the numbers $3,4$, for any $1\leq k \leq 4$ there is one of them divisible by $k$, but $4! = 24 \nmid 12 = 3\cdot 4$.

mavropnevma wrote: Although your way to express it is not altogether correct, it is true that for any $1\leq k \leq n$ at least one term is divisible by $k$. However, this does not guarantee the product of the terms is divisible by $n!$; think about it ... for example, looking at the numbers $3,4$, for any $1\leq k \leq 4$ there is one of them divisible by $k$, but $4! = 24 \nmid 12 = 3\cdot 4$. yes, understood.

Let $n$ consecutive positive integers be $m+1,\dots ,m+n(m\in \mathbb N_{0},n\in \mathbb N)$.Then $\prod_{k=1}^{n}(m+k)=n!\binom{m+n}{n}$ Therefore the proof is completed.$\blacksquare$

lucasxia01 wrote: Thus, the expression is divisible by anything less than $n$ so it remains to prove that it is divisible by $n$. However, this is clear since every residue is taken (by the $n$ consecutive numbers) so we conclude that $n!$ is a factor of the expression. Hello,lucasxia01. You say that if $A$ is divisible by $1\le \forall k\le n$,then $n!|A$,don't you? This is wrong. Counterexample:$A=12, n=4$.$k|12(1\le \forall k\le 4)$,but $4!\nmid 12$. Your sincerely. Takeya.O

Newton binomial

Note that $$\frac{(k+1)(k+2)\cdots(k+n)}{n!}=\binom{n+k}{n}\in \mathbb Z$$