proof
First, define $A_1 \approxeq A_2 \iff ord(A_1) = ord(A_2)$
$\forall \ \text{well-ordered sets}\; A,B \;\text{s.t.} A\cap B=\varnothing$ and $ord(A) = \alpha,\; ord(B)= \beta$,
define $\alpha\beta = ord(A\times B)$ where $A\times B$ denotes anti-lexicographic-order between A and B.
Then take well-ordered set $A', B'$ s.t. $A\cap B=\emptyset\;$ and $ord(A') = \alpha,\; ord(B')= \beta$.
Then since $ord(A') = ord(A)$ and $ord(B') = ord(B)$,
$A' \approxeq A$ and $B \approxeq B'$, which means
$\exists g: A \to A'$ and $\exists h: B \to B'$ s.t.
$\forall x,y \in A, x \le y \Rightarrow g(x) \le g(y)$
and $\forall x,y \in B, x \le y \Rightarrow h(x) \le h(y)$
and $\forall x,y \in A', x \le y \Rightarrow g^{-1}(x) \le g^{-1}(y)$
and $\forall x,y \in B', x \le y \Rightarrow h^{-1}(x) \le h^{-1}(y)$
Now define $f:A \times B \to A' \times B'$ s.t.
$f(x,y) = (g(x), h(y)) \forall x \in A$ and $\forall y \in B$
Then $\forall (x,y)\;$and $(x',y') \in A \times B$
$(x,y) \le (x',y') \Rightarrow f(x,y) \le f(x',y')$ since $g(x) \le g(x')$ and $h(y) \le h(y')$
Similarly,
$\forall (x,y)\;$and $(x',y') \in A' \times B'$,
$(x,y) \le (x',y') \Rightarrow f^{-1}(x,y) \le f^{-1}(x',y')$ since $g^{-1}(x) \le g^{-1}(x')$ and $h^{-1}(y) \le h^{-1}(y')$.
Thus, $A \times B \approxeq A' \times B'$, which menas
$ord(A\times B) = ord(A'\times B')$
Thus, $ord(A'\times B') = \alpha\beta$ Q.E.D.