nt.number theory - On the prime factorisation of g(k) = (k^2)! * prod (j=0, k-1, j!/(j+k)!)

The trick is to simplify as much as possible. Have you noticed that the $n$ never appears alone in your question, but only in the context of $p(n)$, so why don't you just write $p$ for $p(n)$ ? Then, what is sod(j+k, p(n)) ? Notice that if $1leq kleqfrac{p-1}{2}$, then the number $j+k$ (for $jleq k-1$) can only have one digit in base $p$. It's a bit more complicated for $j+(p-k)$; it can have one or two digits. Both $k^2$ and $left(p-kright)^2$ can also have one or two digits, but not more. Things simplify if you proceed further this way.

But it's better not to work with digits at all - digits are chaotic, they tend to split arguments into many cases (because of carries). A more suitable formula is

$vleft(u!,pright)=sumlimits_{l=1}^{infty}leftlfloorfrac{u}{p^l}rightrfloor$, where $p$ is prime and $u$ is a nonnegative integer.

This sum is not really infinite; it's a sum of the form $text{finitely many nonzero addends}+0+0+0+...$ (because $leftlfloorfrac{u}{p^l}rightrfloor=0$ for sufficiently high $l$).

Now let $1leq kleq frac{p-1}{2}$. Then,

$vleft(gleft(kright),pright)=vleft(k^2!cdotprod_{j=0}^{k-1}frac{j!}{left(j+kright)!},pright)$

$=vleft(k^2!,pright)+sumlimits_{j=0}^{k-1}left(vleft(j!,pright)-vleft(left(j+kright)!,pright)right)$

$=sumlimits_{l=1}^{infty}leftlfloorfrac{k^2}{p^l}rightrfloor+sumlimits_{j=0}^{k-1}left(sumlimits_{l=1}^{infty}leftlfloorfrac{j}{p^l}rightrfloor -sumlimits_{l=1}^{infty}leftlfloorfrac{j+k}{p^l}rightrfloorright)$.

Now, the sum $sumlimits_{l=1}^{infty}leftlfloorfrac{k^2}{p^l}rightrfloor$ stops at $l=1$, because $k^2 < p^2$ and thus all the terms for $lgeq 2$ are zero. The sums $sumlimits_{l=1}^{infty}leftlfloorfrac{j}{p^l}rightrfloor $ and $sumlimits_{l=1}^{infty}leftlfloorfrac{j+k}{p^l}rightrfloor$ have not even one nonzero addend, i. e. we can forget about them.

Computing $vleft(gleft(j+p-kright),pright)$ goes the same way, but this time we cannot forget about the sum $sumlimits_{l=1}^{infty}leftlfloorfrac{j+p-k}{p^l}rightrfloor$; it may be nonzero, although it can only have one nonzero term.

I'll leave you the further computations. At the end you will have to prove that

$leftlfloor frac{k^2}{p}rightrfloor = leftlfloor frac{(p-k)^2}{p}rightrfloor - sumlimits_{j=1}^{p-k-1}leftlfloor frac{j+p-k}{p} rightrfloor$.

Now, noticing that $leftlfloor frac{j+p-k}{p} rightrfloor$ can only take the values $0$ and $1$ (because $frac{j+p-k}{p} < 2$), and actually is $1$ if $jgeq k$ and $0$ otherwise, we see that $sumlimits_{j=1}^{p-k-1}leftlfloor frac{j+p-k}{p} rightrfloor$ is simply the number of all $jinleftlbrace 1,2,...,p-k-1rightrbrace$ satisfying $jgeq k$. In other words, $sumlimits_{j=1}^{p-k-1}leftlfloor frac{j+p-k}{p} rightrfloor = left(p-k-1right)-k+1=p-2k$. So it remains to prove that

$leftlfloor frac{k^2}{p}rightrfloor = leftlfloor frac{(p-k)^2}{p}rightrfloor - left(p-2kright)$.

But $leftlfloor frac{(p-k)^2}{p}rightrfloor - left(p-2kright)=leftlfloor frac{(p-k)^2}{p} - left(p-2kright)rightrfloor = leftlfloor frac{k^2}{p}rightrfloor$,

qed.