[proofplan]
We use the product order on $P\times Q$ and first identify exactly which comparable pairs are covers. A cover in the product changes one coordinate by a cover in its factor and leaves the other coordinate fixed. This cover description shows that the sum of the two factor ranks is zero on minimal elements and increases by one along each cover, hence is a rank function. The rank [generating function identity](/theorems/2212) then follows by expanding the finite sum over $P\times Q$ as a double sum and factoring it.
[/proofplan]
[step:Identify the covers in the product poset]
Write $a\lessdot_P b$ to mean that $b$ covers $a$ in $P$, and write $u\lessdot_Q v$ to mean that $v$ covers $u$ in $Q$. We claim that, for $(p,r),(p',r')\in P\times Q$,
\begin{align*}
(p,r)\lessdot_{P\times Q}(p',r')
\end{align*}
if and only if exactly one of the following two alternatives holds:
\begin{align*}
p\lessdot_P p' \text{ and } r=r',
\end{align*}
or
\begin{align*}
p=p' \text{ and } r\lessdot_Q r'.
\end{align*}
Suppose first that $(p,r)\lessdot_{P\times Q}(p',r')$. Since $(p,r)<(p',r')$, we have $p\le_P p'$ and $r\le_Q r'$, with at least one strict inequality. If both inequalities were strict, then
\begin{align*}
(p,r)<(p',r)<(p',r')
\end{align*}
would be a strict intermediate element of $P\times Q$, contradicting that $(p',r')$ covers $(p,r)$. Hence exactly one coordinate changes. If $r=r'$, then $p<_P p'$. If there were $s\in P$ with $p<_P s<_P p'$, then
\begin{align*}
(p,r)<(s,r)<(p',r)
\end{align*}
would be an intermediate element in $P\times Q$, again impossible. Thus $p\lessdot_P p'$. If instead $p=p'$, then $r<_Q r'$. If there were $s\in Q$ with $r<_Q s<_Q r'$, then
\begin{align*}
(p,r)<(p,s)<(p,r')
\end{align*}
would be an intermediate element in $P\times Q$, again impossible. Thus $r\lessdot_Q r'$.
Conversely, suppose $p\lessdot_P p'$ and $r=r'$. If
\begin{align*}
(p,r)\le (a,b)\le (p',r)
\end{align*}
for some $(a,b)\in P\times Q$, then $p\le_P a\le_P p'$ and $r\le_Q b\le_Q r$. Hence $b=r$, and since $p\lessdot_P p'$, either $a=p$ or $a=p'$. Therefore $(a,b)$ is either $(p,r)$ or $(p',r)$, so $(p,r)\lessdot_{P\times Q}(p',r)$.
Now suppose $p=p'$ and $r\lessdot_Q r'$. If
\begin{align*}
(p,r)\le (a,b)\le (p,r')
\end{align*}
for some $(a,b)\in P\times Q$, then $p\le_P a\le_P p$ and $r\le_Q b\le_Q r'$. Hence $a=p$, and since $r\lessdot_Q r'$, either $b=r$ or $b=r'$. Therefore $(a,b)$ is either $(p,r)$ or $(p,r')$, so $(p,r)\lessdot_{P\times Q}(p,r')$.
[guided]
The point of this step is to rule out covers that move in both coordinates at once. In the product order, an inequality
\begin{align*}
(p,r)\le (p',r')
\end{align*}
means simultaneously $p\le_P p'$ and $r\le_Q r'$. Thus, if both coordinates strictly increase, then we can pause after increasing only the first coordinate:
\begin{align*}
(p,r)<(p',r)<(p',r').
\end{align*}
This element lies strictly between the endpoints, so the endpoint cannot cover the starting point.
Therefore a cover in $P\times Q$ must change exactly one coordinate. Suppose the changing coordinate is the $P$-coordinate, so $r=r'$ and $p<_P p'$. We must still check that $p'$ covers $p$ inside $P$. If not, there would be an element $s\in P$ satisfying
\begin{align*}
p<_P s<_P p'.
\end{align*}
Then
\begin{align*}
(p,r)<(s,r)<(p',r)
\end{align*}
would be a strict intermediate element in the product, contradicting the cover relation. Hence $p\lessdot_P p'$. If the changing coordinate is the $Q$-coordinate, then $p=p'$ and $r<_Q r'$. If some $s\in Q$ satisfied
\begin{align*}
r<_Q s<_Q r',
\end{align*}
then
\begin{align*}
(p,r)<(p,s)<(p,r')
\end{align*}
would be a strict intermediate element in the product. This contradicts the cover relation, so $r\lessdot_Q r'$.
For the converse, assume $p\lessdot_P p'$ and $r=r'$. To prove that $(p',r)$ covers $(p,r)$ in the product, take any element $(a,b)\in P\times Q$ satisfying
\begin{align*}
(p,r)\le (a,b)\le (p',r).
\end{align*}
The product order gives $p\le_P a\le_P p'$ and $r\le_Q b\le_Q r$. The second pair of inequalities forces $b=r$. Since $p\lessdot_P p'$, the first pair forces $a=p$ or $a=p'$. Hence the only elements between $(p,r)$ and $(p',r)$ are the endpoints themselves, so $(p,r)\lessdot_{P\times Q}(p',r)$.
Now assume $p=p'$ and $r\lessdot_Q r'$. To prove that $(p,r')$ covers $(p,r)$ in the product, take any element $(a,b)\in P\times Q$ satisfying
\begin{align*}
(p,r)\le (a,b)\le (p,r').
\end{align*}
The product order gives $p\le_P a\le_P p$ and $r\le_Q b\le_Q r'$. The first pair of inequalities forces $a=p$. Since $r\lessdot_Q r'$, the second pair forces $b=r$ or $b=r'$. Hence the only elements between $(p,r)$ and $(p,r')$ are the endpoints themselves, so $(p,r)\lessdot_{P\times Q}(p,r')$.
[/guided]
[/step]
[step:Verify that the sum of the factor ranks is a rank function]
Define $\rho_{P\times Q}:P\times Q\to \mathbb N\cup\{0\}$ by
\begin{align*}
\rho_{P\times Q}(p,r)=\rho_P(p)+\rho_Q(r).
\end{align*}
Since $P$ and $Q$ are finite, $P\times Q$ is finite. We first verify that $\le_{P\times Q}$ is a partial order. Reflexivity follows because $p\le_P p$ and $r\le_Q r$ for every $(p,r)\in P\times Q$. If $(p,r)\le_{P\times Q}(p',r')$ and $(p',r')\le_{P\times Q}(p,r)$, then antisymmetry in $P$ gives $p=p'$ and antisymmetry in $Q$ gives $r=r'$, so $(p,r)=(p',r')$. If $(p,r)\le_{P\times Q}(p',r')$ and $(p',r')\le_{P\times Q}(p'',r'')$, then transitivity in $P$ and $Q$ gives $p\le_P p''$ and $r\le_Q r''$, so $(p,r)\le_{P\times Q}(p'',r'')$. Hence $(P\times Q,\le_{P\times Q})$ is a finite poset. We use the defining characterization of the rank function on a finite graded poset: it vanishes on minimal elements and increases by one along each cover relation. Thus it remains to verify these two properties for $\rho_{P\times Q}$.
Let $(p,r)\in P\times Q$ be minimal. Then $p$ is minimal in $P$ and $r$ is minimal in $Q$, because any smaller coordinate would give a smaller element in the product. Since $\rho_P$ and $\rho_Q$ are rank functions, $\rho_P(p)=0$ and $\rho_Q(r)=0$, so
\begin{align*}
\rho_{P\times Q}(p,r)=0.
\end{align*}
Now let $(p,r)\lessdot_{P\times Q}(p',r')$. By the cover characterization, either $p\lessdot_P p'$ and $r=r'$, or $p=p'$ and $r\lessdot_Q r'$. In the first case,
\begin{align*}
\rho_{P\times Q}(p',r')=\rho_P(p')+\rho_Q(r)=\rho_P(p)+1+\rho_Q(r)=\rho_{P\times Q}(p,r)+1.
\end{align*}
In the second case,
\begin{align*}
\rho_{P\times Q}(p',r')=\rho_P(p)+\rho_Q(r')=\rho_P(p)+\rho_Q(r)+1=\rho_{P\times Q}(p,r)+1.
\end{align*}
Thus $\rho_{P\times Q}$ is zero on minimal elements and increases by one along every cover, so $P\times Q$ is graded with rank function $\rho_{P\times Q}$.
[/step]
[step:Factor the finite rank generating sum]
Let $t$ be the polynomial variable in the rank generating functions. Using the definition of $\rho_{P\times Q}$ and the finiteness of $P$ and $Q$, we compute
\begin{align*}
R_{P\times Q}(t)=\sum_{(p,r)\in P\times Q}t^{\rho_{P\times Q}(p,r)}.
\end{align*}
Substituting the rank formula gives
\begin{align*}
R_{P\times Q}(t)=\sum_{(p,r)\in P\times Q}t^{\rho_P(p)+\rho_Q(r)}.
\end{align*}
Rewriting the finite sum over $P\times Q$ as an iterated finite sum,
\begin{align*}
R_{P\times Q}(t)=\sum_{p\in P}\sum_{r\in Q}t^{\rho_P(p)+\rho_Q(r)}.
\end{align*}
Since powers of the same indeterminate multiply by adding exponents,
\begin{align*}
t^{\rho_P(p)+\rho_Q(r)}=t^{\rho_P(p)}t^{\rho_Q(r)}.
\end{align*}
Therefore
\begin{align*}
R_{P\times Q}(t)=\sum_{p\in P}\sum_{r\in Q}t^{\rho_P(p)}t^{\rho_Q(r)}.
\end{align*}
For each fixed $p\in P$, the factor $t^{\rho_P(p)}$ is independent of $r$, so
\begin{align*}
R_{P\times Q}(t)=\sum_{p\in P}t^{\rho_P(p)}\sum_{r\in Q}t^{\rho_Q(r)}.
\end{align*}
The inner sum is $R_Q(t)$, and the outer sum is $R_P(t)$. Hence
\begin{align*}
R_{P\times Q}(t)=R_P(t)R_Q(t).
\end{align*}
This proves the multiplicativity of rank generating functions.
[/step]
