Let $k$ be a fixed positive integer and $\operatorname{val}_p$ be the $p$-adic valuation on $\mathbb{Z}_p$. Let $n$ be a natural number. My question is finding the following limit:
$$\lim_{n \rightarrow \infty}\frac{n^k}{p^{\operatorname{val}_p(n!)}}.$$
I guess that this limit should be zero; because I guess that $\operatorname{val}_p(n!)$ is like $\dfrac{n}{p-1}$ for large $n$ (I know for sure that $\operatorname{val}_p(n!)$ is certainly less than $\dfrac{n}{p-1}$) and I apply the classical L.Hopital's rule in calculus (although its not relevant to the $p$-adic case) to the limit $\lim_{x \rightarrow \infty} \dfrac{x^k}{p^{x/(p-1)}}$ and by L.Hopital's rule, taking repeated derivatives, I get that $\lim_{x \rightarrow \infty} \dfrac{x^k}{p^{x/(p-1)}}=0$ which gives me the feeling that $\lim_{n \rightarrow \infty} \dfrac{n^k}{p^{\operatorname{val}_p(n!)}}=0$.
Thanks in advance for help and for explanations
The $p$-adic valuation of $n!$ is classically known; see e.g. Artin-Tate's "Class Field Theory", beginning of chapter 12. Writing $n = a_0 + a_1 p + ... + a_r p^r$, where the coefficients $a_j$ are integers between $0$ and $p-1$, one has $v_p (n!) = (n-s_n)/(p-1)$ , where $s_n$ is the sum of the coefficients $a_j$. Since $v_p(n)= t+v_p(a_t)$, where $t$ is the smallest integer such that $a_t \neq 0$, the $p$-adic valuation of your quotient « behaves like » $kt – (n-s_n)/(p-1)$. It follows that your limit is actually $0$.