The product of $n$ consecutive integers is divisible by $ n!$ (without using the properties of binomial coefficients) How can we prove, without using the properties of binomial coefficients, the product of $n$ consecutive integers is divisible by $n$ factorial?
 A: Since this statement is equivalent to the fact that binomial coefficients are integral it is a bit tricky to make precise the notion of a proof that does not "use properties of binomial coefficients". I will interpret this proviso to mean that the proof is not essentially combinatorial, i.e. it does not employ properties that are immediate from the combinatorial interpretation of binomial coefficients.
In my post here you'll find a purely arithmetical proof that $\rm\: n!\: $ divides the product of $\rm\:n\:$ consecutive integers. The proof shows how to rewrite the associated fraction as a product of fractions whose denominators are all coprime to any given prime $\rm\:p\:$. This implies that no primes divide the denominator (when written in lowest terms), hence the fraction is an integer.
The key property that lies at the heart of this proof is that, among all products of $\rm\: n\:$ consecutive integers, $\rm\ n!\ $ has the least possible power of $\rm\:p\:$ dividing it - for all primes $\rm\:p\:$. Thus $\rm\ n!\ $ divides every product of $\rm\:n\:$ consecutive integers, since it has a smaller power of every prime divisor. 
A: Let's define 
$$P(n,k) = \frac{\prod\limits_{i=1}^{n}(k+i)}{n!}.$$
If $n$ is zero, the product is zero. If $k$ is zero, the product is $n!$. Both are divisible by $n!$.
If $k < -n < 0$ then,
$$P(n,k) = (-1)^nP(n,-k-n).$$
As such it suffices to prove this for non-negative $k$.
Induction: 
Base case: 
If $n+k = 1$ then $P(0,1) = 0 \in \mathbb{N}$ and $P(1,0) = 1 \in \mathbb{N}$.
Inductive step: 
Suppose that $P(n,k) \in \mathbb{N}$ for all $n+k = z.$
Then for any $P(n,k)$ with $n+k = z+1$,
$$\begin{gather} P(n,k) = \frac{\prod\limits_{i=1}^{n}(k+i)}{n!}\\
=(k+n)\frac{\prod\limits_{i=1}^{n-1}(k+i)}{n!} \\
=k\frac{\prod\limits_{i=1}^{n-1}(k+i)}{n!} + n\frac{\prod\limits_{i=1}^{n-1}(k+i)}{n!}\\
=\frac{\prod\limits_{i=1}^{n}((k-1)+i)}{n!} + \frac{\prod\limits_{i=1}^{n-1}(k+i)}{(n-1)!}\\
=P(n,k-1) + P(n-1,k).
\end{gather}
$$
Both terms are natural numbers and hence $P(n,k) \in \mathbb{N}$. This completes the induction and the proof.
A: There is an alternate proof at the following link which uses the result about the highest power of a prime dividing a factorial (scroll down for the proof)
http://2000clicks.com/MathHelp/BasicFactorialConsecutiveIntegerProducts.aspx
A: You could argue by induction on $n$. The result is obvious if $n=1$.
For the inductive step, assume that $(n-1)!$ divides any product of $n-1$ consecutive integers. Now consider products of $n$ consecutive integers. 
Say your product is $P(k)=(k+1)(k+2)\dots(k+n)$. First, you may assume $k\ge0$: If one of the factors $k+j$ is 0, then $P(k)=0$ and obviously $n!$ divides it. If $k+n=-t-1<0$, this is the same as $$(-1)^n(t+1)(t+2)\dots(t+n)=(-1)^nP(t),$$ and $t\ge0$. 
Now argue by induction on $k$. The result clearly holds if $k=0$, since $P(0)=n!$.
Suppose $n!|P(k)$. Then $P(k+1)=(k+2)(k+3)\dots(k+n+1)$. Split the last factor as $(k+1)+n$. We have $$P(k+1)=[(k+2)\dots(k+n)](k+1)+[(k+2)\dots(k+n)]n.$$ The first summand is $P(k)$, which is divisible by $n!$ by the inductive assumption on $k$. The second summand is $n$ times a product of $n-1$ consecutive integers, thus $n$ times a multiple of $(n-1)!$, by the inductive assumption on $n$. Clearly, $n$ times a multiple of $(n-1)!$ is a multiple of $n!$, and the sum of two multiples of $n!$ is a multiple of $n!$, so $n!|P(k+1)$.
We are done by induction.
A: My approach would be to show that for any set  of $n$ consecutive numbers, working $\pmod n$, we can see that: $$\forall t\in1,...,n; \exists r\in0,...,n-1:t|an+r$$
In this sense, $k=an+b$ for some $a,b$, as we offset the $\pmod n$.
This is trivially true because one in every $t$ consecutive integers is divisible by $t$.
So, if we worked with $n!=1\cdot2\cdot\ldots\cdot n$ we match with $\underbrace{(k+1)(k+2)\ldots(k+n)}_{n \text{ consecutive numbers}}$, and note that for each $t\leq n$ there is a multiple of $t$ in each subsection of the product of size $t$, and thus the same is true across the whole product.
A: Almost self contained proof without induction. (Let $p$ be an arbitrary prime)
We will denote by $\nu_p$ the $p$-adic valuation and let $s_p\left(\sum_{j=0}^{n}{a_jp^j}\right)=\sum_{j=0}^{n}{a_j}$ denote the sum of digits in base $p$.
We claim that $$\nu_p(n!)=\frac{n-s_p(n)}{p-1}$$ This may be directly verified by subsituting $n=\sum_{j=1}^{k}{a_jp^j}$ and using Legendre's formula: $$\nu_p(n!)=\sum_{j=1}^{\infty}{\left\lfloor \frac{n}{p^j} \right\rfloor}$$ This inequality can be easily shown: $$s_p(a+b)\leq s_p(a)+s_p(b)$$ Therefore,$$\nu_p\left(\binom{n}{k}\right)=\nu_p\left(\frac{n!}{k!(n-k)!}\right)=\frac{s_p(k)+s_p(n-k)-s_p(n)}{p-1}\geq 0$$ hence we have proved that $$\binom{n}{k}=\frac{n(n-1)\ldots(n-k+1)}{k!}\in \mathbb{Z}$$ [Using :"If $x\in\mathbb{Q}$ such that $\nu_p(x)\geq 0$ for all primes $p$ then $x\in \mathbb{Z}$"]
