Prove that
length(P x Q) = length(P) + length(Q)
where P, Q are ordered sets of finite length.
Any ideas of where to start...
If $\displaystyle a_1,\dots,a_n$ and $\displaystyle b_1,\dots,b_m$ are chains in P and Q, respectively, then $\displaystyle (a_1,b_1),(a_2,b_1),\dots,(a_n,b_1),$$\displaystyle (a_n,b_2),\dots,(a_n,b_m)$ is a chain in P x Q, so length(P x Q) >= length(P) + length(Q). Conversely, if you have a chain $\displaystyle (a_1,b_1)>\dots>(a_n,b_n)$ in P x Q, then for each $\displaystyle i$ you have $\displaystyle a_{i+1}>a_i$ or $\displaystyle b_{i+1}>b_i$ (or both). So, you can construct chains in P and Q whose total length is >= n. This means that length(P) + length(Q) >= length(P x Q).