the largest number $y$ in which $(x!)^{x+y}|(x^2)!$

Question

Since the multiplication of $n$ consecutive integers is divided by $n!$, then $(n!)^n|(n^2)!$ with $n$ is a positive integer.

Are there any formula of the function $y=f(x)$ that shows the largest number $y$ in which $(x!)^{x+y}|(x^2)!$ with $x>1$. If so, for a positive integer $y$, are there infinitely many $x$ such that $(x!)^{x+y}|((x^2)!)$ (with $x, y, n$ are all positive integers) ? If not, what are the conditions of $m$ so that there are infinitely many $x$ such that $(x!)^{x+m}|((x^2)!)$ ? (with $m$ is a positive integer and $m \leq y$)

Answer 1

Take the numbers from $(k-1)n+1$ to $kn$. The first $n-1$ of them are a multiple of $(n-1)!$ so the full $n$ are a product of $k(n!)$.
So the product of all $n^2$ numbers gives $n$ lots of $n!$ and the $k$ give an $n+1$th.
That means that $y(x)\ge1$ for all $x$

Answer 2

Consider a prime $p \le x$. The $p$-adic order of $x!$ is $\nu_p(x!) = \sum_{k \ge 1} \lfloor x/p^k \rfloor$. Since $x \lfloor x/p^k \rfloor \le \lfloor x^2/p^k\rfloor$ for all $p$, we have $(x!)^x \mid (x^2)!$.
The largest $y$ is the minimum over all such $p$ of $\lfloor \dfrac{\nu_p((x^2)!)}{\nu_p(x!)} - x \rfloor$.

Hmmm. It appears to match OEIS sequence A187279 (verified up to $n=10000$).