Proving $r!$ divides the product of r succesive positive integers I have to prove the following theorem:

Prove that the product of $r$ consecutive positive integers in divisible by $r!$

I am having a hard time getting a generalization down for the full set of real numbers, if I start from 1 and work up to r, I have the following:
$$r!k=\prod_{i=1}^{r}n_i$$
Can easily prove the base case of this, (n=1), and then go in to prove:
$$(r+1)!k=\prod_{i=1}^{r+1}n_i$$
Expand that out and get:
$$(r+1)r!k=n(n+1)(n+2).....(n+r)(n+r+1)$$
Can say that the product of the first $r$ elements in equal to $r!k$ by our base case. Leaving using with:
$$(r+1)k=(n+r+1)$$
Not sure where I can go from here, n is the integer that we start at, so how can I get it to work out to be equal to our induction hypothesis?
 A: You can do this by simultaneous induction on $r$ and $n$.  Note that
$$\begin{align}
(n+1)\cdots(n+r)&=(n+1)\cdots(n+r-1)n\quad+\quad(n+1)\cdots(n+r-1)r\\
&=((n-1)+1)\cdots((n-1)+r)\quad+\quad(n+1)\cdots(n+(r-1))r
\end{align}$$
(I inserted a little extra space around the central plus signs to make the key pieces easier to see.)  By induction on $n$, $r!$ divides $((n-1)+1)\cdots((n-1)+r)$, and by induction on $r$, $(r-1)!$ divides $(n+1)\cdots(n+(r-1))$, hence $r!$ divides $(n+1)\cdots(n+(r-1))r$.  
(Please note, I'm glossing over all the fine points of getting the inductions started.)
A: $\newcommand{\+}{^{\dagger}}%
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$$
n\pars{n + 1}\ldots\pars{n + r - 1}={\pars{n + r -1}! \over \pars{n - 1}!}
={\pars{n + r -1}! \over \pars{n - 1}!r!}\,r! = {n + r - 1 \choose r}r!
$$

$$
{n\pars{n + 1}\ldots\pars{n + r - 1} \over r!} = {n + r - 1 \choose r}
\quad\mbox{which is an integer !!!} 
$$

