TITLE: The product of $n$ consecutive integers is divisible by $n$ factorial
QUESTION [46 upvotes]: 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.

REPLY [6 votes]: 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