Direct proof that $n!$ divides $(n+1)(n+2)\cdots(2n)$

Question

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 $\binom{2n}{n}=\frac{(2n)!}{n!n!}$.

Answer 1

We present a proof that uses just basic properties of divisibility. Note that $(2n)!=n!(n+1)(n+2)\cdots (2n)$. The result will follow if we prove the following stronger lemma.

Lemma: The product of any $n$ consecutive positive integers is divisible by $n!$.

Proof of Lemma: The proof is by a double induction. We suppose that the product of any $n$ consecutive positive integers is divisible by $n!$, and show that the product of any $n+1$ consecutive positive integers is divisible by $(n+1)!$.

Consider the product $$P(w)=(w+1)(w+2)\cdots(w+n+1).$$ We show by induction on $w$ that $P(w)$ is divisible by $(n+1)!$ for any non-negative integer $w$. The result is trivially true if $w=0$. We show that if the result holds for $w$, it holds for $w+1$. We have $$P(w+1)=(w+2)(w+3)\cdots (w+n+2).$$ By bringing out common factors, we observe that $$P(w+1)-P(w)=(w+2)(w+3)\cdots(w+n+1)[(w+n+2)-(w+1)]=(w+2)(w+3)\cdots(w+n+1)[n+1].\tag{1}$$ The product $(w+2)(w+3)\cdots(w+n+1)$ is the product of $n$ consecutive positive integers, so by the outer induction hypothesis it is divisible by $n!$. It follows from (1) that $P(w+1)-P(w)$ is divisible by $(n+1)!$. But by the inner induction hypothesis, $P(w)$ is divisible by $(n+1)!$. It follows that $P(w+1)$ is divisible by $(n+1)!$. This completes the proof.

Remark: This is really an answer to a recent previous question asking for a "divisibility" proof of the fact that $(n!)^2$ divides $(2n)!$. That question (which I cannot find) was closed as a duplicate of a question of which it was not a duplicate, and which is referred to in the OP.

Answer 2

I am not really interested in reading other threads but here is an alternate proof.

Take a positive integer $d \le n$. Then suppose $n+1 \equiv a \pmod d$. Then the numbers $n+1, n+2, \cdots, 2n$ cycle through atleast one complete residue class $\pmod d$ since the next residue is just $+1$ the previous. Then by CRT, we are done.

Answer 3

Let $p$ be a prime and $k$ an arbitrary integer such that $p^k\le n$, and let $b$ be the largest positive integer such that $p^k\cdot b\le n$. Then $p^k(b+1)\gt n$ and $p^k\cdot 2b\le 2n$. By this, every multiple of $p^k$ from $1$ to $b$ appears in the interval $[1,n]$, and similarly for the $b+1$ to $2b$ multiples which appear in $[n+1,2n]$, and thus the number of times a factor $p^k$ appears in $n!$ is less than or equal to the number of times it appears in $(2n)!\over n!$. But $p$ and $k$ were arbitrarily chosen as $p^k\in[1,n]$, so this applies to every power of every prime that appears in the interval $[1,n]$. Therefore

$$n!\mid \frac{(2n)!}{n!}$$

Answer 4

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.