Find the interval of convergence of the power series $\sum_{n=0}^\infty \frac{(np)!(x^n)}{(n!)^p}$ where p is a natural number.

Question

I tried to use the ratio test to find the radius of convergence of this power series and evaluated $R$ to be the following: $$R=\frac{1}{p^p}.$$ This would imply that the series converges absolutely in the interval ($\frac{-1}{p^p}$,$\frac{1}{p^p}$). I then put $x=\frac{-1}{p^p}$ to test the convergence at the endpoints. This would result into the following series: $$\sum_{n=0}^\infty \frac{(-1)^n(np)!}{(n!)^p\cdot p^{np}}.$$ I then tried to use the absolute convergence test and tried to test the convergence of the series $$\sum_{n=0}^\infty \frac{(np)!}{(n!)^p\cdot p^{np}}.$$ I know that this series diverges for $p=1$ because when p=1, the series becomes $$\sum_{n=0}^\infty1$$ and this series diverges as $$\lim_{n\to\infty}1=1$$ which is not zero and thus by nth term test the series diverges for p=1. However I cannot think of a way to show in general that this series diverges for any natural number $p$. I tried to use the ratio test but this results in $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$=1 and thus the test fails. As I have not been taught the Raabe's test or Stirling formula, I am looking for a solution that does not use these concepts.