If $f = a_0 + a_1t + \cdots\in\mathbb{Z}_p[[t]]$ has $p\mid a_0$ but $p\nmid f$, could $1/f$ have bounded denominators in $\mathbb{Q}_p[[t]]$? Let $f\in\mathbb{Z}_p[[t]]$ be a power series, say
$$f = a_0 + a_1t + a_2t^2 + \cdots$$
Suppose $p\nmid f$ (ie, there exists some $a_i$ which is a unit in $\mathbb{Z}_p$), but suppose $p\mid a_0$.
Is it possible for the coefficients of $\frac{1}{f}$ as an element of $\mathbb{Q}_p[[t]]$ to have $p$-adic valuation bounded from below? (e.g., there exists a $b\in\mathbb{Z}$ such that all coefficients of $1/f$ have valuation $\ge b$)
Here the $p$-adic valuation of $p^n$ is $n$.
EDIT: The most basic example: $f = p - t$ has inverse $\frac{1}{p} + \frac{1}{p^2}t + \cdots$. Playing around on wolframalpha seems to suggest that the in general the inverses of such $f$ have unbounded denominators, though I don't have a proof.
 A: Note that we need $a_0 \neq 0$ for $1/f$ to be in $\mathbb{Q}_p[[t]]$ and not just $\mathbb{Q}_p((t))$.
Let's appeal, perhaps unnecessarily, to the $p$-adic Weierstrass preparation theorem, which gives
$$
f(t) =  u(t) \cdot g(t),
$$
where $u(t)$ is a unit and $g(t)$ is a degree $n$ polynomial which is equivalent to $t^n$ mod $p$ (i.e. all the non-leading terms are divisible by $p$). It is clear that $n$ is the smallest integer such that the $n$th coefficient of $f$ is a unit, so the hypothesis of the problem is that $n > 0$.
Now as $u$ is a unit, its leading coefficient is a unit, and so is the leading coefficient of $\frac{1}{u} \in \mathbb{Z}_p[[t]]$. It follows that $\frac{1}{f}$ has  coefficients bounded below if and only if $\frac{1}{g}$ does. This reduces us to the case of distinguished polynomials.
Let $g(t) = t^n + pg_0(t)$ is a distinguished polynomial and suppose $h(t) \in p^{-m}\mathbb{Z}_p[[t]] \setminus p^{-m+1}\mathbb{Z}_p[[t]]$ satisfies $g(t)h(t) = 1$. (We are assuming that $m$ exists for the sake of contradiction and we may as well choose the smallest such $m$.)
Then we have the equation $p^mg(t)h(t) = p^m$ in $\mathbb{Z}_p[[t]]$. This gives
$$
p^mt^nh(t) + p^{m+1}g_0(t)h(t) = p^m.
$$
Mod $p$ we have $t^nh(t)p^m = 0$, contradicting the assumption on $h(t)$. There must be no such $m$.
