First of all, note that we need to prove that \[p! \cdot (p^2)! \cdots (p^p)! \mid p \cdot \left(p^2+p^3+\dots+p^p\right)!.\]If it were not for the factor $p!$ on the LHS, this would be obvious by multinomial coefficients. But of course it suffices to prove that each prime $q$ divides the LHS less often than the RHS, and the multinomial coefficient argument shows that this is true for $q>p$. Indeed, it still works for $q=p$ since we have the factor $p$ on the RHS (and indeed this is precisely what we need it for). Now by Legendre's formula, it clearly suffices to prove that \[\left\lfloor \frac{p^2+\dots+p^p}{q}\right\rfloor-\left\lfloor \frac{p^2}{q}\right\rfloor-\dots-\left\lfloor \frac{p^p}{q}\right\rfloor \ge \left\lfloor \frac{p}{q}\right\rfloor\]for all primes or prime powers $q<p$. (Here again, the LHS is trivially non-negative and the RHS comes from the additional factor $p!$.) But note that we can write the LHS as a telescoping sum \[\left(\left\lfloor \frac{p^2+p^3}{q}\right\rfloor-\left\lfloor \frac{p^2}{q}\right\rfloor-\left\lfloor \frac{p^3}{q}\right\rfloor\right)+\left(\left\lfloor \frac{p^2+p^3+p^4}{q}\right\rfloor-\left\lfloor \frac{p^2+p^3}{q}\right\rfloor-\left\lfloor \frac{p^4}{q}\right\rfloor\right)+\dots+\left(\left\lfloor \frac{p^2+p^3+\dots+p^p}{q}\right\rfloor-\left\lfloor \frac{p^2+\dots+p^{p-1}}{q}\right\rfloor-\left\lfloor \frac{p^p}{q}\right\rfloor\right)\]Here each of the $p-2$ summands is either $0$ or $1$. But note that since $q<p$ it will clearly be $1$ whenever $q \mid p^2+p^3+\dots+p^k$. Now if $p \equiv 1 \pmod{q}$, this happens whenever $k \equiv 1 \pmod{q}$ and hence exactly $\frac{p-1}{q}=\left\lfloor \frac{p}{q}\right\rfloor$ many times. Else, it certainly happens for $k \equiv 1 \pmod{\varphi(q)}$ and hence at least $\left\lfloor \frac{p-1}{\varphi(q)}\right\rfloor \ge \left\lfloor \frac{p}{q}\right\rfloor$ many times.

$\nu_q$ chasing

We are asked to prove that the number $A=\frac{p\cdot \left(p^2\cdot\frac{p^{p-1}-1}{p-1}\right)!}{p!\cdot (p^2)! \cdot (p^3)! \cdot \ldots \cdot (p^{p-1})! \cdot (p^p)!}$ is an integer. Write $A=\frac{k}{\ell}$ with $\gcd(k,\ell)=1$. Denote $n=p^2\cdot\frac{p^{p-1}-1}{p-1}$. Note that $$A = \frac{1}{p^s}\binom{n}{p-1, p^2-1, \ldots, p^p-1}$$for some integer $s$ (which happens to be equal to $(p-1)(p+2)/2$ if I counted correctly, but the actual value of $s$ is irrelevant). Therefore $\ell$ is a power of $p$. Moreover $$A = \frac{1}{(p-1)!}\binom{n}{p^2,p^3,\ldots,p^p}.$$This shows that $p\nmid\ell$. Hence $\ell$ is a power of $p$ which is not divisible by $p$, so $\ell=1$ and $A$ is an integer.