The product of $n$ consecutive integers is divisible by $n$ factorial How can we prove that the product of $n$ consecutive integers is divisible by $n$ factorial?

Note: In this subsequent question and the comments here the OP has clarified that he seeks a proof that "does not use the properties of binomial coefficients". Please post answers in said newer thread so that this incorrectly-posed question may be closed as a duplicate.

 A: A clearer version of NurdinTakenov's proof. I prefer Knuth's notation, and falling factorials are nicer to work with:
$\begin{equation*}
m^{\underline{k}} = m (m - 1) \ldots (m - k + 1)
\end{equation*}$
First:
$\begin{align*}
(m + 1)^{\underline{k}} - m^{\underline{k}}
  &= (m + 1) \cdot m^{\underline{k - 1}} 
      - m^{\underline{k - 1}} \cdot (m - k + 1) \\
  &= k \cdot m^{\underline{k - 1}}
\end{align*}$
So:
$\begin{align*}
\sum_{0 \le r \le m - 1} r^{\underline{k}}
   &= \frac{m^{\underline{k + 1}}}{k + 1}
\end{align*}$
Now the proof by induction over $k$ goes through easily:
Base: If $k = 0$, we have that $0! \mid m^{\underline{0}}$, which is just $1 \mid 1$.
Induction: Assume $k! \mid m^{\underline{k}}$ for all $m$. Then:
$\begin{align*}
  m^{\underline{k + 1}} 
    &= (k + 1) \sum_{0 \le r \le m - 1} r^{\underline{k}}
\end{align*}$
By induction, each term of the sum is divisible by $k!$, so the right hand side is divisible by $(k + 1) k! = (k + 1)!$.
If $k > m$, then one of the factors in $m^{\underline{k}}$ is zero, and the result is trivial.
If $m$ is negative, it is clear that $m^{\underline{k}} = (-1)^k
 \lvert m \rvert^{\underline{k}}$, extending the above so it covers this case too.
A: You might be interested in this blog post of Timothy Gowers:
http://gowers.wordpress.com/2010/09/18/are-these-the-same-proof/
A: This answer completely formalizes the argument of Nurdin Takenov in a manner sufficient to easily be expressed in an automated theorem prover such as PVS. Note that this proof uses strong induction on the sum m+k to avoid any nasty double inductions, and is explicit about all assumptions on the arguments: 
DEFINITION: *P*roduct of k consecutive posints starting at m  (m>=1, k>=1)
i.e. P(m,k) ==def== m...(m+k-1)
LEMMA:  P(m,k)  =  k*P(m,k-1)      +  P(m-1,k)  if m>=2 and k>=2
PROOF:  P(m,k)  =  m...(m+k-1)
    =  m...(m+k-2)[ k + (m-1) ]

    =  k*(m)...(m+k-2)  + (m-1)...(m+k-2)

    =  k*P(m,k-1)      +  P(m-1,k)  QED

THEOREM: Product of k consecutive posints starting with m is divisible by k factorial
i.e.  k! | P(m,k)
PROOF (by strong induction on all sums m+k <= n):
(i) BASIS: If n = 2 then clearly m=k=1 and we have k! = 1! clearly divides P(m,k) = 1
(ii) INDUCTION STEP: Assume k! | P(m,k) for all m+k<=n. Now to show that k! | P(m,k) for all m+k <= n+1
If m=1 we are done since P(1,k) = 1...k = k! and if k=1 then k! = 1! clearly divides P(m,k). So in the remainder
we may assume that m >= 2 and k >= 2.  Also if m+k<=n we are done vacuously, so consider only that m+k = n+1.
By the lemma we have P(m,k) = k*P(m,k-1) + P(m-1,k) so by the Induction hypothesis we have   (k-1)! | P(m,k-1)
and thus also k! | k*P(m,k-1)  and also by the Induction hypothesis k! | P(m-1,k) and finally k! | P(m,k) QED
A: This is almost immediate from the fact that the binomial coefficient $$\binom{k+n}{k}$$ is an integer. Just write the product $(k+1) \cdots (k+n)$ accordingly and you'll have your answer.
A: The identity below shows that the problem is equivalent to the fact that binomial coefficients are integral - for which various proofs are known, e.g. using their recursion, or their well-known combinatorial  interpretation, or their minimality in terms of prime divisors - see this prior question
$$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{n\:(n-1)\quad\quad\:\cdots\:\quad\quad 1\quad\quad}$$
A: Let us prove that $m^{(k)}=m(m+1)...(m+k-1)$ is divided by $k!$ for all integer $m$.
Induction by $k$.
$k=1$: Every integer $m$ is divided by $1$
$k\to k+1$: 


*

*induction by $m$: $m=0$: $0^{k+1}=0$
is divided by $(k+1)!$
$m\to m+1$: $(m+1)^{(k+1)}=(m+1)(m+2)...(m+k+1)$
$=(k+1)(m+1)...(m+k)+m^{(k+1)}=(k+1)(m+1)^{(k)}+m^{(k+1)}$
and first term is divided by $(k+1)\cdot k!=(k+1)!$ because of induction by k and the second term is divided by $(k+1)!$ because of induction by $m$
the same works for $m\to m-1$
Update: Oops, essentially the same proof found in the thread mentioned in this answer.
A: For a given prime $p$, the maximum number of times which $p$ can divide $n!$ is
$$\sum_{k=1}^\infty \left[{n\over p^k}\right]$$, where $[x]$ is the floor function
(to get this result, think about the number of multiples of $p^k$ which do not exceed $n$, and the fact that $p^k$ is a multiple of $p^i$ for each $i\le k$).
(Note that the summation above is actually finite.)
Then, the maximum number of times which $p$ can divide $(m+1)\cdots(m+n)=(m+n)!/m!$ is
$$\sum_{k=1}^\infty \left[{m+n\over p^k}\right]-\left[{m\over p^k}\right].$$
Since $[a]+[b]\le[a+b]$, $[(m+n)/p^k]-[m/p^k]\ge[n/p^k]$, so the above is
$$\ge\sum_{k=1}^\infty \left[{n\over p^k}\right],$$
which is actually the maximum number of times which $p$ can divide $n!$.
This is true for all prime $p$, so we get
$$n!\mid(m+1)\cdots(m+n).$$
