Let $p$ be a prime, $x \in \mathbb{Z}_p$. I say that if $x \in \mathbb{Z}_p$ then $\text{val}_p(x) \ge 0$ as follows:
Let $x = a_0 + a_1 p + a_2 p^2 + \dots$. Then, $\text{val}_p(x) \ge \text{min}(\text{val}_p(a_0), \text{val}_p(a_1), \dots ) \ge 0 $.
Now, I think that $\text{val}_p(x) \ge 0 \implies x \in \mathbb{Z}_p$ but could not prove it. I start by assuming $\text{val}_p(x) \ge 0 $ and $x\notin \mathbb{Z}_p$. Since $x\in \mathbb{Q}_p $, say $$x= \dots + a_{-k} \cfrac{1}{p^k} + \dots$$ for some $a_{-k} \in \{1,\dots,p-1\} $. Then, we can write
$$x = \cfrac{1}{p^k}(\dots + a_0 + \dots)$$ and we have $\text{val}_p(x) = \text{val}_p(\cfrac{1}{p^k}) \text{val}_p(\dots + a_0 + \dots) = -k + \text{val}_p(\dots + a_0 + \dots) \ge 0$.
However, I stuck here. So, can we say
$\text{val}_p(x) \ge 0 \implies x \in \mathbb{Z}_p$
or not?
Let $x \in \mathbb{Q}_p$. That means we can write $$x = \sum_{k \geq -m} a_kp^k,$$ where $\text{val}_p(x) = -m$. Now assume that $\text{val}_p(x) \geq 0$. That means $x$ has to be of the form $$x = \sum_{k \geq 0} a_kp^k,$$ which just means it is an element of $\mathbb{Z}_p$.