I've been studying $p$-adic fields and convergence of series in them and I've been working on a proof that $ lim_{n \rightarrow\infty} |n!|_{p} = 0$ in any $p$-adic field. I came up with the following proof and I was just wondering if it was correct.
Let $\epsilon > 0$. Since $n! \in \mathbb{N}$, we have $n! = p^{t}r$, where $r > 0$ and $(p,r)=1$. Let $n > \frac{r}{\epsilon}$. Then,
\begin{equation*} n! = p^{t}r>n>\frac{r}{\epsilon}\implies p^{t} > \frac{1}{\epsilon} \implies p^{-t} < \epsilon. \end{equation*}
Since $|n!|_{p} = |p^{t}r|_{p}=p^{-t}$, we have $|n!|_{p} = p^{-t} < \epsilon$, completing the proof.
I was also trying to show the same limit is true for $n^{2}$ and it seems to me that I could use the exact same method, since $n^{2} = p^{t}r$ is also true. Does this mean that increasing sequences of natural numbers always converge in the $p$-adic fields, since each of them can be expressed in this way?
Hint: For which $n$ is $n!$ divisible by $p$? For all (!) $n \ge ...$
For which $n$ is $n!$ divisible by $p^2$? For all (!) $n \ge ...$
(If you don't see it immediately, just literally write $n!$ down up to, say, $n=15$, and check divisibility for $p=2$, then for $p=3, p=5, p=7$.)
What about $p^3, p^4, ..., p^k$?
(And notice that e.g. for the sequence $a_n := n^2$, there is no $N$ such that $p$ divides all $a_n$ for $n \ge N$; there's always some which are not divisible by $p$, let alone higher powers of $p$.)