Consider the [Vandermonde's determinant](https://en.wikipedia.org/wiki/Vandermonde_matrix) computed by
$$V(x\_1,\dots,x\_m):=\det(x\_j^{i-1})\_{i,j=1}^m=\prod\_{1\leq i<j\leq m}(x\_i-x\_j).$$
The number of [plane partitions](https://en.wikipedia.org/wiki/Plane_partition) in an $n\times m\times m$ box (MacMahon) is given by the neat product formula
$$\prod\_{i=1}^n\prod\_{j=1}^m\prod\_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$


**Notation.** Let $\mathcal{F}\_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j\_1,\dots,j\_m\}$ for $\mathbf{J}\in\mathcal{F}\_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}\_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.


I would like to ask:


**QUESTION.** Is this expansion true? A combinatorial proof is desired, if possible.
$$\prod\_{i=1}^n\prod\_{j=1}^m\prod\_{k=1}^m\frac{i+j+k-1}{i+j+k-2}
=\sum\_{\mathbf{J}\in\mathcal{F}\_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$


**Remark.** Observe the cute fact $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer (let's make Fedor's comment explicit).


**Proof.** Replacing $\mathbf{x}=(x\_1,\dots,x\_m)$ by $\mathbf{J}=(j\_1,\dots,j\_m)$ into
$$\det\left[\binom{x\_j}i\right]
=\frac1{\prod\_{i<j}(j-i)}\det\left[\prod\_{k=0}^{i-1}(x\_j-k)\right]
=\frac{\det\left[x\_j^{i-1}\right]}{\prod\_{i<j}(j-i)}
=\prod\_{1\leq i \lt j\leq m} \frac{x\_j-x\_i}{j-i}.$$
results in integer entries, and thus integer determinant $\frac{V(\mathbf{J})}{V(\mathbf{I})}$. $\qquad\square$



Plane partitions as sums of determinants
I haven't worked out the details, but $V(\mathbf{J})/V(\mathbf{I})$ is
the principal specialization of a Schur function. Then $\left(
\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2$ corresponds to a pair of
SSYT (semistandard Young tableaux), which can be merged into a plane
partition as in EC2, proof of Theorem 7.20.1. Summing over
$\mathbf{J}$ should give all plane partitions fitting in an $n\times
m\times m$ box.


