The $n$th triangular number is defined as $T_2(n) = n(n+1)/2$, and there is an interesting product rule for triangular numbers: $$T_2(mn) = T_2(m)\,T_2(n) + T_2(m-1)\,T_2(n-1).$$
The tetrahedral numbers $T_3(n) = n(n+1)(n+2)/6$ satisfy a similar product rule: $$T_3(mn) = T_3(m)\,T_3(n) + 4 T_3(m-1)\,T_3(n-1) + T_3(m-2)\,T_3(n-2).$$
Can these product rules be generalized to higher dimensions? Are there analogous formulas for $T_k(mn)$, where $$T_k(n) = \frac1{k!} n(n+1)(n+2)\cdots(n+k-1)\ ?$$
On
The n'th K'th-Simplex number is (2 = triangle, 3 = tetrahedral, etc...)
$$T_k{(n)} = \frac{1}{k!} \frac{(n+k-1)!}{(n - 1)!} = \begin{pmatrix} n + k - 1 \\ k\end{pmatrix}$$
It appears that without exploiting the above fact there are multiple product rules that can be derived:
One possible strategy to derive product rules is to use Induction
Note that $$T_k(n) = \frac{n + k - 1}{k} T_{k-1} (n) $$
Then notice
$$T_2(n) - T_2(n-1) = n$$ $$T_3(n) - 2 T_3(n-1) + T_3(n-2) = n$$ And in general
$$ \sum_{i = 0}^{k-1} \left[ (-1)^i \begin{pmatrix} k -1 \\ i \end{pmatrix} T_k(n- i)\right] = n$$
And as a corollary to this
$$T_{k+1}(m) - T_{k+1}(m-1) = T_{k}(m)$$
So now let us attempt to derive a general product rule for $T_k(mn)$ assuming we know the product rule for $T_{k-1}(mn)$ as
$$ T_{k-1} = P(T_{k-1}(m), T_{k-1}(n), T_{k-1}(m-1), T_{k-1}(n-1) ... )$$
We immediately find that each argument $T_{k-1}(w)$ of P can be re-written as $T_{k}(w) - T_{k}(w-1)$ and therefore $P$ can be rewritten as function of terms of the form $T_k$ furthermore:
$$ T_k(mn) = \frac{mn + k - 1}{k}T_{k-1}(mn)$$
$$ T_k(mn) = \frac{mn + k - 1}{k}P$$
Now using the above summation formulas (we conclude)
$$ T_k(mn) = \frac{\left( \sum_{i = 0}^{k-1} \left[ (-1)^i \begin{pmatrix} k -1 \\ i \end{pmatrix} T_k(m- i)\right] \right) \left( \sum_{i = 0}^{k-1} \left[ (-1)^i \begin{pmatrix} k -1 \\ i \end{pmatrix} T_k(n- i)\right] \right) + k - 1}{k} P$$
Thus giving us a messy BUT working general rule for deriving product identities given a previous product identity $P$. Convert each term in this identity into the higher $T_k$ form and then proceed to compute:
$$frac{\left( \sum_{i = 0}^{k-1} \left[ (-1)^i \begin{pmatrix} k -1 \\ i \end{pmatrix} T_k(m- i)\right] \right) \left( \sum_{i = 0}^{k-1} \left[ (-1)^i \begin{pmatrix} k -1 \\ i \end{pmatrix} T_k(n- i)\right] \right) + k - 1}{k}$$
multiply those two and a new product identity has been found
using this we attempt to derive the triangular number identity from just the normal numbers
$$T_1(mn) = T_1(m) T_1(n)$$
Therefore
$$T_1(mn) = (T_2(m) - T_2(m-1))(T_2(n) - T_2(n-1))$$
Now
$$T_2(mn) = \frac{mn + 1}{2}T_2(mn)$$
$$m = T_2(m) - T_2(m-1)$$ $$n = T_2(n) - T_2(n-1)$$
Therefore
$$T_2(mn) = \frac{(T_2(m) - T_2(m-1))(T_2(n) - T_2(n-1))+1}{2}(T_2(m) - T_2(m-1))(T_2(n) - T_2(n-1)) $$
This is and further derivations appear to be non-trivially different from the product rules you have come across. That is unless one actually simplifies both product rules it isn't obvious how to simplify one into the other
By denoting with $(n)_k$ the falling Pochhammer symbol $$(n)_k = n\cdot(n-1)\cdot\ldots\cdot(n-k+1)$$ we have: $$ (mn+1)_2 = \frac{1}{2}\left((m+1)_2(n+1)_2+(m)_2(n)_2\right)\tag{1}$$ that leads to the first identity. Since: $$ (mn+2) = \alpha(n+2)(m+2)+\beta(n-1)(m-1) $$ is solved by $\alpha=\frac{1}{3}$ and $\beta=\frac{2}{3}$ and $$ (mn+2) = \gamma(n+1)(m+1)+\delta(n-2)(m-2)$$ is solved by $\gamma=\frac{2}{3}$ and $\delta=\frac{1}{3}$, multiplying both sides of $(1)$ by $(mn+2)$ we get: $$ (mn+2)_3 = \frac{1}{6}\left((m+2)_3(n+2)_3+4(m+1)_3(n+1)_3+(m)_3(n)_3\right)\tag{2}$$ that leads to the second identity.
Trying to reproduce the same argument, we may look for $\alpha$ and $\beta$ such that: $$(mn+3)=\alpha(m+3)(n+3)+\beta(m-1)(n-1),$$ so equating the coefficients of $mn$ we get $\alpha+\beta=1$ and equating the coefficients of $(m+n)$ we get $3\alpha-\beta=0$, from which $\alpha=\frac{1}{4},\beta=\frac{3}{4}$. The constant terms match since $\frac{9}{4}+\frac{3}{4}=3$. So we go on and try to find $\gamma,\delta$ such that: $$(mn+3)=\gamma(m+2)(n+2)+\delta(m-2)(n-2).$$ However, by choosing $\gamma=\delta=\frac{1}{2}$ the constant term does not match. Ouch. This gives that the formula for pentatope numbers depends on tetrahedral numbers, too. For istance, by multiplying both sides of $(2)$ by $(mn+3)$ we get: $$\begin{array}{rcl}(mn+3)_4 &=& \frac{1}{24}\left((m+3)_4(n+3)_4+11(m+2)_4(n+2)_4\right. \\&+&\left.11(m+1)_4(n+1)_4+(m)_4(n)_4\color{red}{-16(m+1)_3(n+1)_3}\right).\end{array}\tag{3}$$
If we don't care introducing further terms, we can simply go on as we did for producing $(3)$ from $(2)$.