Show that the product of any $m$ consecutive positive integers is divisible by $m!$ 
Show that the product of any $m$ consecutive positive integers is divisible by $m!$.

Note that we have that $\frac{n!}{m!(n-m)!} \in \mathbb{Z}$ for $0 \leq m \leq n$.
$\require{cancel} \frac{n!}{m!(n-m)!} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)(n-m)!}{m!(n-m)!} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)\cancel{(n-m)!}}{m!\cancel{(n-m)!}} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)}{m!}$ where $n\cdot(n-1)\cdot\cdot\cdot(n-m+1)$ are $m$ consecutive integers divisible by $m!$
Of course, I'm using the fact that I know the combinations formula is an integer but given that, is the proof this simple? Or did I make an error here. Appreciate help!
 A: If $p$ is a prime and $n$ is a nonzero integer, then let us define $v_p(n)$ as the largest nonnegative integer $k$ such that $p^k\ |\ n$.
Legendre's (or de Polignac's) formula says that $v_p(n!) = \sum_{i=1}^{\infty}\lfloor \frac{n}{p^i} \rfloor$.
Enough to show that  $v_p(n!) \geq  v_p(m!) + v_p((n-m)!)$ for every prime number $p$.
Using the fact that $\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x 
 + y\rfloor$, we can write  $\lfloor \frac{m}{p^i} \rfloor + \lfloor \frac{n-m}{p^i} \rfloor \leq \lfloor \frac{n}{p^i}\rfloor\ \forall\ i$
So we get that $v_p(m!) + v_p((n-m)!) \leq v_p(n!)$, and so $\frac{n!}{m!(n-m)!} \in \mathbb{Z}$ 
Using the fact that the formula is an integer as well as the proof given in the question, we have that the product of any $m$ consecutive integers is divisible by $m!$
A: $n(n-1)...(n-m+1)$ is the number of choices of $m$ distinct objects in a set of $n$ elements  (there exists $n$ way to choose $1$ object, if an object is chosen, there exists $n-1$ way to chose the second, so $n(n-1)$ way to choose $2$ objects and recursively,...) (first operation: choose $m$ distinct objects in a set of $n$ objects), given a set $M$ of $m$ objects $m!$ is the number of bijection of $M$, this implies that each choice appears $m!$ times (with a different order ) in the first operation.
A: Both answers posted so far assume that $n \geq m$, which is not given in the problem statement.
However, this assumption can be justified as a WLOG assumption. Indeed, fix a nonnegative integer $m$. We want to prove that
\begin{align}
m! \mid n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right)
\qquad \text{for each } n \in \mathbb{Z} .
\label{darij1.eq.1}
\tag{1}
\end{align}
Assume that we have proven that
\begin{align}
m! \mid n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right)
\qquad \text{for each integer } n \geq m .
\label{darij1.eq.2}
\tag{2}
\end{align}
(The currently existing answers prove this.) Now, let $n \in \mathbb{Z}$ be arbitrary. Since $m!$ is a positive integer, the arithmetic sequence $\left(n, n+m!, n+2m!, n+3m!, n+4m!, \ldots\right)$ of integers grows unboundedly, and thus eventually surpasses the number $m$. In other words, there exists some nonnegative integer $k$ such that $n+km! \geq m$. Consider this $k$. Let $N = n+km!$. Then, $N$ is an integer satisfying $N = n+km! \equiv n \mod m!$ and $N = n+km! \geq m$. Hence, \eqref{darij1.eq.2} (applied to $N$ instead of $n$) yields
\begin{align}
m! \mid N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) .
\end{align}
In other words,
\begin{align}
N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) \equiv 0 \mod m! .
\end{align}
But we have $N \equiv n \mod m!$. Thus, $N - i \equiv n - i \mod m!$ for each $i \in \left\{0,1,2,\ldots,m-1\right\}$. Multiplying all these $m$ congruences together, we find
\begin{align}
N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) \equiv
n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right) \mod m! .
\end{align}
Hence,
\begin{align}
& n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right) \\
& \equiv
N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) \\
& \equiv 0 \mod m! .
\end{align}
In other words, $m! \mid n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right)$. Thus, \eqref{darij1.eq.1} is proven.
