Well in a book i am reading it is given that you can also prove this by showing that
Every prime factor is contained in $(n+r)!$ as often at least as it is contained in $n!r!$.
How does this prove the proposition, i am not getting the meaning of this.
Help appreciated.
On
Try some test cases, and look at the remainders of the r consecutive digits. Do you see a pattern? You may also want to consider the pigeonhole principle too.
On
You consider each prime to r separately.
For example, if r were 10, then the primes to consider are 2, 3, 5, and 7.
For 5 and 7, there are at least 2 and 1 multiples of 5 and 7. So these are ok.
For 2, there are 5 multiples of 2, so we can cancel these out of both sides. Likewise we see that there are two multiples of 4, and a multiple of eight, which amount to cross out 3 2's.
A similar proof can be done for 9, alternately, a strip of 9 contains all nine modulos, so lining up the multiples of 9 give all multiples of 3 as well.
Therefore, any 10 consecutive numbers is a multiple of 10!
On
Another way is to note that $\binom{n+r}{r}$ counts the number of ways to choose $r$ distinct objects from $n+r$ objects and hence must be an integer...
On
If the numbers have different signs, $0$ is among them, and the proposition is trivial.
If not, we can WOLOG suppose that they are positive.
For any given $n\in\Bbb N$ and a prime $p$, let $\nu_p(n)$ the greatest integer $k$ such that $p^k$ that divides $n$ (it can be $0$, of course).
Then $$\begin{align} \nu_p\left(\prod_{k=1}^r(n+k)\right)&=\nu_p\left(\frac{(n+r)!}{n!}\right)\\ &=\sum_{\alpha=1}^\infty\left\lfloor\frac{n+r}{p^\alpha}\right\rfloor- \sum_{\alpha=1}^\infty\left\lfloor\frac{n}{p^\alpha}\right\rfloor \end{align}$$
It suffices to show that for each $n$, $p$ and $\alpha$, $$\left\lfloor\frac{n+r}{p^\alpha}\right\rfloor-\left\lfloor\frac{n}{p^\alpha}\right\rfloor\geq\left\lfloor\frac{r}{p^\alpha}\right\rfloor$$
For this, write $\left\lfloor\frac{n+r}{p^\alpha}\right\rfloor=\frac{n+r}{p^\alpha}-\epsilon_1$ and $\left\lfloor\frac{n}{p^\alpha}\right\rfloor=\frac{n}{p^\alpha}-\epsilon_2$, where $0\leq\epsilon_1,\epsilon_2<1$. Then, $$\left\lfloor\frac{n+r}{p^\alpha}\right\rfloor-\left\lfloor\frac{n}{p^\alpha}\right\rfloor=\frac r{p^\alpha}-(\epsilon_1-\epsilon_2)$$
Since $\epsilon_1-\epsilon_2<1$, this concludes the proof.
Note: The difference $\epsilon_1-\epsilon_2$ is always between $-1$ and $1$. If it's positive or zero, we have $$\left\lfloor\frac{n+r}{p^\alpha}\right\rfloor-\left\lfloor\frac{n}{p^\alpha}\right\rfloor=\left\lfloor\frac{r}{p^\alpha}\right\rfloor$$ and if it is negative $$\left\lfloor\frac{n+r}{p^\alpha}\right\rfloor-\left\lfloor\frac{n}{p^\alpha}\right\rfloor=\left\lfloor\frac{r}{p^\alpha}\right\rfloor+1$$
On
Define rising factorial powers by: $$ x^{\overline{m}} = x (x + 1) \ldots (x + m - 1) $$ It is easy to prove that: $$ (n + 1)^{\overline{m}} - n^{\overline{m}} = m \cdot n^{\overline{m - 1}} $$ Turning this around gives: $$ \sum_{1 \le k \le n} k^{\overline{m}} = \frac{n^{\overline{m + 1}}}{m + 1} $$ Now we can prove that $n^{\overline{r}}$ is divisible by $r!$ by induction on $r$.
But $\frac{(n + r)!}{n!} = (n + 1)^{\overline{r}}$, and apply the above. Thus $n!r!$ divides $(n + r)!$.
The product of some $r$ consecutive integers can be represented as $$\overbrace{(n+r)(n+r-1)\cdots(n+1)}^{r\mathrm{\ consective\ integers}}=\frac{(n+r)!}{n!}$$ where $n$ is the number one less than the smallest of the consecutive integers. Now, if it is true that primes in $(n+r)!$ appear just as frequently or more as in $n!r!$, then you are saying that for some integer $k$ (likely big) that $(n+r)!=k\cdot n!r!$. So your product of $n$ consecutive integers is $$\frac{(n+r)!}{n!}=\frac{k\cdot n!r!}{n!}=k\cdot r!$$ and is therefore divisible by $r!$.