Proof using induction: $r!$ divides the product of $r$ consecutive integers 
Prove by induction that the product $$n(n+1)...(n+r-1)$$ of any $r$ consecutive numbers is divisible by $r!$.

In the inductive case i've$$n(n+1)...(n+k-1)(n+k)$$
I was not able to find a way to prove this is divisible by (k+1)!.
I try to use distributive property in (n+k).
 A: 
The product of $r$ consecutive numbers is divisible by $r$.

Proof: Okay, I've just remodeled my answer. For $r=1$ it is obvious. 
Let it be true for $r=k$. For $r=k+1$, we will induct by the integer $m$, while showing that $(k+1)! | m(m+1)...(m+k)$. So suppose that $(k+1)! | m(m+1)...(m+k)$.We want to show that $(k+1)! | (m+1)(m+2)...(m+k+1)$.
How we do this is as follows: define $m_k = (m)(m+1)...(m+k)$. Then,
$$
\begin{equation}
\begin{split}
& (m+1)_k =  (m+1)_k -m_k + m_k\\
& =(m+1)...(m+k)(m+k+1-m) + m_k\\
& =(k+1)(m+1)_{k-1} + m_{k}
\end{split}
\end{equation}
$$
Now, note that $k! | (m+1)_{k-1}$ by induction on $k$, hence $(k+1)! |(k+1)(m+1)_{k-1}$. On the other hand, by induction on $m$, $(k+1)! | m_{k}$. Hence, $(k+1)!$ divides their sum, which is $(m+1)_k$. Now, we are done by induction on $m$, and then by induction on $k$.
A: Without induction: The expression
$$
n(n+1)\cdots(n+r-1)
$$
counts the number of words of length $r$ with all distinct letters, from an alphabet of $n+r-1$ letters.
But this is also equal to the number of sets of $r$ letters, times $r!$ since there are $r!$ ways of arranging the set into a word. Thus, the above expression is divisible by $r!$.
A: If $r=1$ the theorem obviously holds true.
Let $r$ be the least natural number for which the theorem does not hold true and $a$ the least natural such that $R_a=(a+2)(a+3)\cdots (a+r+1)$ is not divisble by $r!$.
Obviously, $R_a>r!$ and from now we suppose that any product of $r-1$ numbers is a multiple of $(r-1)!$
We can see that  $R_a-R_{a-1}=$
$(a+2)(a+3)\cdots (a+r)[a+r+1-(a+1)]=(a+2)(a+3)\cdots (a+r)\cdot r$.    
It is obvious that $(a+2)(a+3)\cdots (a+r)$ is a product of $r-1$ numbers which means it is divisible by $(r-1)!$ (from induction hypothesis)so, multiplying by $r$ , it becomes a multiple of $r!$.
Hence $r!\mid R_a-R_{a-1}$ and $r!\mid R_{a-1}$ so, $r!\mid R_a$ which contradicts the definition of $a$.
