Evaluate the limit without using Stirling Formula

You can use the product development of sine instead of the Stirling formula.

$\displaystyle\lim_{n \to \infty} p\frac{(n!)^p p^{np}}{(np)!n^{(p-1)/2}} = p\prod\limits_{k=1}^{p-1}\Gamma\left(\frac{k}{p}\right)= p\sqrt{\prod\limits_{k=1}^{p-1}\Gamma\left(\frac{k}{p}\right)\Gamma\left(1-\frac{k}{p}\right)}=$

$\hspace{3.8cm}\displaystyle = p\sqrt{\prod\limits_{k=1}^{p-1}\frac{\pi}{\sin(\frac{\pi k}{p})}} = p\sqrt{\frac{\pi^{p-1}}{2^{1-p}p}}=\sqrt{p(2\pi)^{p-1}}$

Note:

$\displaystyle \prod\limits_{k=1}^{p-1}\sin\left(x+\frac{\pi k}{p}\right) = \frac{1}{(i2)^{p-1}} \prod\limits_{k=1}^{p-1}\left( e^{i(x+\frac{\pi k}{p})}- e^{-i(x+\frac{\pi k}{p})}\right)$

$\displaystyle = \frac{1}{(i2)^{p-1}} \prod\limits_{k=1}^{p-1} e^{i\frac{\pi k}{p}} \prod\limits_{k=1}^{p-1}\left( e^{ix}- e^{-ix-i2\frac{\pi k}{p}}\right)$

$\displaystyle = \frac{1}{(i2)^{p-1}} e^{i\frac{\pi(p-1)}{2}}\frac{e^{ipx}-e^{-ipx}}{e^{ix}-e^{-ix}} = 2^{1-p}\frac{\sin(px)}{\sin x} ~~(\,\to p2^{1-p}\,\,$ for $\,x\to 0\,)$

From the de Moivre - Laplace theorem, or at least one particular form of it, we have that $$ {an \choose n} \left(\frac{1}{a}\right)^n \left(\frac{a-1}{a}\right)^{(a-1)n} \sim \frac{1}{\sqrt{2 \pi an \frac{1}{a} \left(\frac{a-1}{a}\right)}} \tag{1} $$ as $n \to \infty$, for every positive integer $a$. (The symbol "$\sim$" denotes that the limit of the quotient of left and right sides is 1.) The most direct way to prove this is by using Stirling's formula, but you can prove it without Stirling's formula by more general probability theory. See for example Theorem 7 in Tao's notes on variants of the central limit theorem.

Your equation (as expressed in your edit) is $$ \lim_{n \to \infty} a \frac{(n!)^a a^{an}}{(an)! n^{(a-1)/2}} = a^{1/2} (2\pi)^{(a-1)/2}, $$ which can also be written as $$ \frac{(an)!}{(n!)^a} \sim \sqrt{a} \frac{a^{an}}{(2 \pi n)^{(a-1)/2}}. $$ This is equivalent to equation (1), by using $$ \frac{\frac{(an)!}{(n!)^a}}{\frac{((a-1)n)!}{(n!)^{(a-1)}}} = {an \choose n} $$ and induction on $a$.