$\mathbb Z^n$ under dictionary order.

How could I start proving this:

Let $n \geq 2$. Does the set $\mathbb{Z^n}$ under dictionary order satisfy the greatest lower bound property? If so, prove it. If not, provide (and prove) a counterexample.

-
Before I write anything, which way do you think it will go? Yes, or no? –  Brian M. Scott Oct 10 '12 at 6:46
You should also explain what precise property you're talking about. –  Chris Eagle Oct 10 '12 at 13:49

Consider the set $S = \{(-n, 0, \ldots, 0): n \in \Bbb N\}$. For any $(a_1, \ldots, a_n) \in \Bbb Z^n$, there is some $n \in \Bbb N$ so that $-n < a_1$; hence, $(a_1, \ldots, a_n)$ cannot be the lower bound. As it was an arbitrary element of $\Bbb Z^n$, the set $S$ does not have a lower bound.