Let $A$ be a finite, totally-ordered alphabet (set of characters), and let $S$ denote the set of all strings (of any length, possibly infinite) over this alphabet, ordered lexicographically. Lexicographical ordering is defined as follows: for any two strings $s_1,s_2 \in S$, $s_1 < s_2$ iff the first character in $s_1$ that differs from the corresponding character in $s_2$ is the smaller of the two. (The edge case in which one of the strings is a proper prefix of the other is handled by thinking of strings as being terminated by a special character that is smaller than all characters in $A$).
Let $\alpha$ denote the order type of $[|A|] \stackrel{def}{=} \{1, \ldots, |A|\}$, so that $\alpha^\mathbb{N}$ denotes the product $\prod_{i = 1}^{\infty} \alpha$ (i.e. the order type of $[|A|]^{\mathbb{N}}$, the set of all infinite sequences in $[|A|]$, where for any $(x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}} \in [|A|]^{\mathbb{N}}$ we have $(x_n)_{n \in \mathbb{N}} < (y_n)_{n \in \mathbb{N}}$ iff they're not equal and the elements at the last differing position $i$ satisfy $x_i < y_i$).
Would I be right in saying that the order type of $S$ is $\alpha^\mathbb{N}$? Is there any standard notation for this kind of order type? Also, what would happen if $A$ were not finite, but say infinitely countable, would the order be $\omega^\mathbb{N}$ by extension?
Start with $[0,1]$, and split each dyadic rational in $(0,1)$ into two adjacent points; call the resulting linear order $X$.
When $|A|=2$, the set of infinite sequences over $A$ ordered lexicographically is order-isomorphic to $X$: essentially we’re just looking at the binary expansions of numbers in $[0,1]$ but not identifying the two expansions for dyadic rationals. Thus, $\langle 0,1,1,1,\ldots\rangle$ immediately precedes $\langle 1,0,0,0,\ldots\rangle$. As long as $2\le|A|<\omega$, the set of infinite sequences over $A$ ordered lexicographically will be order-isomorphic to a set like $X$, but built from a countable dense subset of $(0,1)$ different from $D$.
The lexicographic order on $S$ is rather more complicated. In the example above, for instance, the immediate successor of $\langle 0,1,1,1,\ldots\rangle$ is now $\langle 1\rangle$, which is immediately followed by $\langle 1,0\rangle$, and that by $\langle 1,0,0\rangle$, and so on, so that a whole sequence converging monotonically to $\langle 1,0,0,0,\ldots\rangle$ is inserted between $\langle 0,1,1,1,\ldots\rangle$ and $\langle 1,0,0,0,\ldots\rangle$.
More generally, let $\langle d_0,d_1,\ldots,d_n\rangle$ be a finite sequence such that $d_n\ne 0$; then $\langle d_0,d_1,\ldots,d_n\rangle$ is the immediate successor of $\langle d_0,d_1,\ldots,d_n-1,a,a,a,\ldots\rangle$, where $a=\max A$. It is immediately followed by $\langle d_0,d_1,\ldots,d_n,0\rangle$, which is immediately followed by $\langle d_0,d_1,\ldots,d_n,0,0\rangle$, and so on. Finally, the order has an initial sequence of order type $\omega$, namely, $\langle 0\rangle$, $\langle 0,0\rangle$, $\langle 0,0,0\rangle$, and so on, converging up to the infinite zero sequence.