Product of k Consecutive Numbers is Divisible by k!

Suppose $P$ is the product of $k$ consecutive integers, where $k \geq 1$.
Then, $k!$ divides $P$ .


Proof 1 -

Let $P$ be the product of $k$ consecutive integers from $n+1$ up to $n+k$, so that we have:

$$ P = (n+1)(n+2)...(n+k) = \frac{(n+k)!}{n!} $$

Now, we know that the power of any prime $p$ in $k!$ is given by the following finite sum:

$$ \sum\limits_{m \geq 1}^{}{\left\lfloor \frac{k}{p^m} \right\rfloor} $$

Therefore, the power of any prime $p$ in the product $P$ is given by:

$$ \sum\limits_{m \geq 1}^{}{\left\lfloor \frac{n+k}{p^m} \right\rfloor}
- \sum\limits_{m \geq 1}^{}{\left\lfloor \frac{n}{p^m} \right\rfloor} $$
Since $ \forall p,q,r \in \mathbb{Z}^{+}: \left\lfloor \frac{p+q}{r} \right\rfloor \geq
\left\lfloor \frac{p}{r} \right\rfloor + \left\lfloor \frac{q}{r} \right\rfloor $, we must have:

$$ \sum\limits_{m \geq 1}^{}{\left\lfloor \frac{n+k}{p^m} \right\rfloor}
- \sum\limits_{m \geq 1}^{}{\left\lfloor \frac{n}{p^m} \right\rfloor}
\geq \sum\limits_{m \geq 1}^{}{\left\lfloor \frac{k}{p^m} \right\rfloor} $$

Thus, we see that, for every prime $p$, the power of $p$ in $k!$ never exceeds the power of $p$ in the product $P$. Hence, $k!$ always divides the product $P$.

Proof 2 -

Let $P$ be the product of $k$ consecutive integers from $n+1$ up to $n+k$. So, we have:
$ \begin{aligned}
\hspace{4cm} P & = (n+1)(n+2)...(n+k) \\
& = \frac{(n+k)!}{n!} \\
& = k!  \frac{(n+k)!}{n!k!} \\
& = k!  \binom{n+k}{k}
\end{aligned} $

Now, the binomial co-efficient $\binom{n+k}{k}$, being the number of ways in which one can choose $k$ items out of a total of $n+k$ distinct items, is known to be integral. Hence, $k!$ must divide the product $P$.