If $A_1, \ldots, A_n$ are well ordered sets, then so is the Cartesian product $\prod_{i=1}^n A_i$ under the dictionary order. Am I right?
Is this finite product also well ordered in the anti-dictionary order?
Now suppose that $J$ is an infinite set of indices, and suppose $\left\{ A_\alpha \right\}_{\alpha \in J}$ is a collection of well ordered sets. Then is the set $$A \colon= \prod_{\alpha \in J} A_\alpha$$ also well ordered in the dictionary order? under the anti-dictionary order?
If so, then how to prove this rigorously?
If not, then how to construct a counter example?
The result does not hold for infinite products.
Here's a counterexample: take each $A_i$ to be $\{0, 1\}$ ordered the usual way, $i\in\mathbb{N}$. Then let $e_i$ be the string with a $1$ in the $i$th place and a $0$ everywhere else; what can you say about $e_i$ versus $e_j$ if $i<j$?
Also, it's worth noting that the dictionary order doesn't really make sense if $J$ isn't well-ordered: how else do you know that "the first place where $F$ and $G$ disagree" even exists, if $J$ isn't well-ordered?
For a concrete example of this, consider $J=\{. . . , -3, -2, -1, 0\}$ and let $A_i=\{0, 1\}$ (ordered as usual) for each $i\in J$. Now consider the sequences $$F=( . . . , 0, 1, 0, 1),\quad G=( . . . , 1, 0, 1, 0).$$ Is $F<G$ or is $G<F$?