Direct proof that $n!$ divides $(n+1)(n+2)\cdots(2n)$ I've recently run across a false direct proof that $n!$ divides $(n+1)(n+2)\cdots (2n)$ here on math.stackexchange. The proof is here prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer (it is the one by user pedja, which got 11 upvotes). The proof is wrong because it claims that one can rewrite $(n+1)\cdots (2n)$ as
$$ (n+1)(n+2)\cdots 2(n-2)(2n-1)(2n) = 2\cdots 2\cdot n!\cdot (2n-1)(2n-3)\cdots (n+1).$$
In other words, it claims that the product of the factors $2n$, $2(n-1)$, $2(n-2)$, $\ldots$, all of which are in $(n+1)\cdots(2n)$, amounts to $2^kn!$, but this is not true since the factors $2m$ under scrutiny do not start from $m=1$ but from values greater than $n$. For instance, for $n=4$, we have $(8)(7)(6)(5)=2\cdot 2\cdot 4\cdot 3\cdot 5\cdot 7$, not $(8)(7)(6)(5)=2\cdot 2\cdot 4!\cdot 5\cdot 7$. This makes me wonder two things:
(1) What is a valid direct proof?
(2) How many wrong proofs do go undetected here? (How many false proofs receive 10+ upvotes?)
NB Not interested in any proof that uses binomial coefficients and/or the relationship $\binom{2n}{n}=\frac{(2n)!}{n!n!}$.

Here is a more direct number theoretical type proof that if $a \ge 0$ and $a_1+a_2+\cdots+a_r = n$ that $\frac{n!}{a_1!a_2! \cdots a_r!}$ is an integer.
This reduces to proving $\sum \left \lfloor \frac{n}{p_i}  \right \rfloor  \ge \sum \left \lfloor \frac{a_1}{p_i}  \right \rfloor  + \sum \left \lfloor \frac{a_1}{p_i} \right \rfloor + \cdots + \sum \left \lfloor \frac{a_r}{p_i}  \right\rfloor $.
Lemma: $\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x+y \rfloor$ if $x,y$ are real numbers. 
This can be easily proved by writing $x,y$ in terms of their integer and fractional parts.
Applying this to the $RHS$ gives $RHS \le \left \lfloor \frac{a_1+a_2+ \cdots + a_r}{p_i} \right \rfloor = \lfloor \frac{n}{p_i} \rfloor$ and doing this over all $i$ and all $p$ gives the desired  result.
