How to prove that $\frac{1000!}{(500!)^2}$ is not divisible by 7? 
How to prove that 
  $$\frac{1000!}{(500!)^2}$$
  is not divisible by 7?

I reduced the above fraction to:
$$\frac{\prod_{k=501}^{1000}k}{500!} $$
But I don't know how to proceed. Please help.
 A: It is a general formula that the number of times a prime $p$ occurs in the prime factorisation of $n!$ is given by
$$
\left[\frac{n}{p}\right] + \left[\frac{n}{p^2}\right] + \left[\frac{n}{p^3}\right] + \cdots
$$
where $[x]$ denotes the greatest integer $≤x$.
In $1000!$ you'd have
$$
\left[\frac{1000}{7}\right] + \left[\frac{1000}{7^2}\right] + \left[\frac{1000}{7^3}\right] + \cdots = 164
$$
$7$'s, and in $500!$ you have
$$
\left[\frac{500}{7}\right] + \left[\frac{500}{7^2}\right] + \left[\frac{500}{7^3}\right] + \cdots = 82
$$
$7$'s. So you'll have a total of $164$ 7's in $(500!)^2$.
The rest follows.
A: You can also overkill it by Kummer's theorem:
Notice that $500$ is $1313$ in base $7$. Since every digit is smaller than $\frac{7}{2}$ there are no carries when adding $1313+1313$ in base $7$.
Therefore $\binom{1000}{500}$ is not a multiple of $7$.

In general we have that a prime $p$ divides $\binom{2n}{n}$ if and only if $n$ has at least one digit larger than or equal to  $p/2$ in its base $p$ expansion.
A: Consider prime factorisations:
$$\lfloor\frac{1000}{7}\rfloor+\lfloor\frac{1000}{7^2}\rfloor+\lfloor\frac{1000}{7^3}\rfloor+...=142+20+2=164$$
Then $$7^{164}|1000!$$
Similarly:
$$\lfloor\frac{500}{7}\rfloor+\lfloor\frac{500}{7^2}\rfloor+\lfloor\frac{500}{7^3}\rfloor+...=71+10+1=82$$
Then $$7^{164}=7^{82\cdot2}=(7^{82})^2|(500!)^2$$
A: This is overkill, but by Lucas's Theorem$$\frac{1000!}{(500!)^2}=\binom{1000}{500} \equiv \binom{2}{1}\binom{6}{3}\binom{2}{1}\binom{6}{3} \equiv 4 \pmod{7}$$
As $1000=2626_{7}$ and $500=1313_{7}$. 
Here is another , more elementary argument, from Legendre's formula,  $$\nu_7(n!) = \frac{n - s_7(n)}{6}$$
So $\nu_7(100!)=164, \nu_7 (50!)=82$. So the $7$ s in the denominator and the numerator cancel out. So we have the desired result. 
A: In $\{1,\dots, 500\}$, there are 71 multiples of 7, 10 multiples of $7^2$ and one multiple of $7^3$.
How many multiples of $7$, $7^2$, $7^3$ do you have in $\{501,\dots, 1000\}$? Same amount.
