Take limit in p-adic integer ring: the difference between $\lim_{n \rightarrow \infty} \alpha^{n!}$ and $\lim_{n \rightarrow \infty} \alpha^{n}$

Question

Let $p$ be a prime, $L / \mathbb{Q}_p$ be a finite field extension. Let $\mathcal{O}_L$ its valuation ring (ring of local integers) in $L$, $\mathfrak{m}_L$ be the unique maximal ideal, and the cardinality of the residue field $\mathbb{F}$ is $q$.

I am hoping to show that for any $\alpha \in \mathcal{O}_L$, $$\lim_{n \rightarrow \infty} \alpha^{n!} = \begin{cases} 0, \quad \alpha \in \mathfrak{m}_L, \\ 1, \quad \alpha \in \mathcal{O}_L^{\times}. \end{cases} $$

Here is the proof that I have read before:

• For $\alpha \in \mathfrak{m}_L$, since $\vert \alpha \vert_p < 1$, the sequence $\{ \vert \alpha \vert_p^{n!} \}$ tends to zero.
• For $\alpha \in \mathcal{O}_L^{\times}$, we count $\# (\mathcal{O}_L / \mathfrak{m}_L^{r+1})^{\times} = q^r(q-1)$ Then $\alpha^{q^r(q-1)} \equiv 1 \bmod{\mathfrak{m}_L^{r+1}}$. Therefore, $$ \lim_{n \rightarrow \infty} \alpha^{n!} = \lim_{r \rightarrow \infty} \alpha^{q^r(q-1)} = 1. $$

My question is: where had I used the property of "$n!$"? In other words, by simply replacing $n!$ by $n$ everywhere in the proof, it seems that I have "proved " $$\lim_{n \rightarrow \infty} \alpha^{n} = \begin{cases} 0, \quad \alpha \in \mathfrak{m}_L, \\ 1, \quad \alpha \in \mathcal{O}_L^{\times}. \end{cases} $$

Hence I must have misunderstood something but I cannot figure it out. It seems that the only place that I do not fully understand is the equality $$ \lim_{n \rightarrow \infty} \alpha^{n!} = \lim_{r \rightarrow \infty} \alpha^{q^r(q-1)}. $$ Could someone help me to explain why this hold or say why $$ \lim_{n \rightarrow \infty} \alpha^{n} = \lim_{r \rightarrow \infty} \alpha^{q^r(q-1)}. $$ may not hold?

Sorry for such a stupid question, and thank you all for commenting and answering!

Answer 1

Define $\nu$ as the largest natural number for which $n!$ is divisible by $p^{\nu-1}(p-1)=\phi(p^\nu)$. Then $|\alpha^{n!}-1|=p^{-\nu}$ in the $p$-adic norm. What happens with $n!$, but not just $n$, is that $\nu$ is both (nonstrictly) monotonically increasing and unbounded, conditions which force $|\alpha^{n!}-1|$ to converge to $0$.