The positive integers under divisibility ordering (i.e. $p\le q$ if $p$ divides $q$) form a distributive lattice, but do they also form a complete lattice?
My argument is that whatever subset you take, for example, the subset of all prime numbers, then meet will be $1$ and join will be $\prod_{p=1}^{\infty }p_{i}$
On
An non-empty ordered set $P$ is a complete lattice if and only if, for every subset $S$ of $P$, the meet (infimum) and the join (supremum) of $S$ both exist.
(Davey, B. A.; Priestley, H. A., Introduction to lattices and order., Cambridge: Cambridge University Press. xii, 298 p. (2002). ZBL1002.06001.)
For our purposes, we can define divisibility to mean that $a|b \iff b=ak$, with $a,b,k \in\mathbb{N}^{*}$.
The positive integers $\mathbb{N}^{*}$ under divisibility ordering ($a\le b \iff a|b$) do not form a complete lattice because there is no upper bound in the case of the subset $S=\mathbb{N}^{*}$ (that is, there $\not\exists \alpha \in\mathbb{N}^{*}$ such that $n|\alpha$ for all $n\in\mathbb{N}^{*}$). This can be proved by contradiction.
Suppose, for the sake of contradiction, that there $\exists \alpha \in \mathbb{N}^{*}$ such that $n|\alpha$ for all $n\in\mathbb{N}^{*}$.
Because the positive integers are closed under multiplication, $2\alpha \in\mathbb{N}^{*}$. Then by definition, $2\alpha|\alpha$. Because $\alpha \not=0$, we can simplify on both sides to get $2|1$, which is false under our definition of divisibility, and as such, we have our contradiction.
The non-negative integers $\mathbb{N}$ with divisibility, however, do form a complete lattice, with a meet of $1$ and a join of $0$. This is because $\forall n \in\mathbb{N}, 1|n$ and $n|0$. Our definition of divisibility is important when we look at $n|0$, especially when $n=0$.
You're right, if you consider the set of all prime numbers then it has no least upper bound.
The same idea show that there is no maximum element for the whole lattice, so it certainly cannot be complete. To see this, suppose for the sake of a contradiction that $N$ is maximum with respect to $\mid$. Write $N = p_1^{n_1}\cdots p_m^{n_m}$ as the prime factorization of $N$, where $p_1,p_2,\ldots,p_n$ are the first $n$ prime numbers. By hypothesis, $p_{n+1} \mid N$, but this is a contradiction.
Note that if we added in $0$ (with the understanding that $a \mid 0$ for every $a$, which is justified if we define our divisibility relation by saying that $n \mid m$ if there exists an integer [not necessarily non-zero] $x$ such that $nx=m$), then we would have a complete lattice. $0$ would be our maximum, and $0$ would also be the join of the set $\{ p \mid \text{$p$ prime}\}$.