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).
On
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 $r!$.
On
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 $a$.
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 $k$.