I was reading the following paper and saw the comment (on page 2) that with p a prime $p \geq 5 $ and $p=\frac{n+2}{2}$ that $p^3\parallel(2n)!$ (where in this case the double pipe notation means that if $p^a\parallel t$ then $p^a\mid t$ but $p^{a+1} \nmid t$)
I am sure that there is some reasonably simple justification for this, but I am not so clever. My justification is quite long winded, and I can't really bring myself to move on without knowing the more simple justification for this fact. Primarily, I was curious if anyone could please provide a more simple answer than mine.
I will provide the way I justified this:
$$ \mu_q(m!)=\sum_{k=1}^{\infty} \left \lfloor \frac{m}{q^k}\right \rfloor $$
is the highest power of q which divides m!, applying this to the above problem
$$ \mu_p((2n)!)=\sum_{k=1}^{\infty} \left \lfloor \frac{2n}{\left(\frac{n+2}{2}\right)^k} \right \rfloor < 2n\sum_{k=1}^{\infty}\frac{2^k}{(n+1)^k}=2n\left(\frac{1}{1-\frac{2}{n+2}}-1\right)=4 $$
So that $\mu_p((2n)!)<4$, and therefore $p^4 \nmid (2n)!$
Now consider only the first term of the sum $\mu_p((2n)!)$,
$$ \left \lfloor \frac{4n}{n+2}\right \rfloor $$
Since $p\ge5$: $n \ge 8$, so taking n to be 8
$$ \left \lfloor \frac{4\times8}{8+2}\right \rfloor=\left \lfloor \frac{32}{10}\right \rfloor=\left \lfloor 3.2 \right \rfloor=3 $$
Now again considering only the first term and keeping in mind that floor does not change the sign here
$$ \frac{d}{dn} \frac{4n}{(n+2)}=\frac{8}{(n+2)^2} $$
So that the first term is an increasing function of n, now since $$3\le \mu_p((2n)!) < 4$$ and $\mu_p((2n)!)$ must be an integer, then $\mu_p((2n)!)=3$. Therefore $$ p^3\parallel(2n)! $$
If I did anything stupid please feel free to let me know. Thank you for reading.
Let $p>5$ be a prime number, and let $n=2p-2$ (which is equivalent to $p=\frac{n+2}2$). Then $p$, $2p$, and $3p$ are all less than $2n=4p-4$, but $4p$ is not. Therefore, $(2n)!$ has exactly three terms that are multiples of $p$. Moreover, since $p>5$, all three terms are less than $p^2$, so each term carries exactly "one power of $p$". Thus, $p^3\parallel(2n)!$ as desired.