In $ℚ[Z]$, by $Z \choose k$ denote the polynomial $${Z \choose k} = \frac{1}{k!}·\prod_{i=0}^{k-1} (Z-i),$$ so that ${Z \choose k}(n) = {n \choose k} = \frac{n!}{k!·(n-k)!}$ in $ℚ$.
Now, in the ring of formal power series $ℚ[Z] [\![ X, Y ]\!]$, define the binomial series of $F ∈ (X,Y)$ as $$(1+F)^Z = \sum_{k=0}^∞ {Z \choose k}F^k.$$ (Or do that specifically for $F = X$, $F = Y$ and $F = X + Y + XY = (1+X)(1+Y) - 1$.)
A natural question is now whether $$(1+X)^Z(1+Y)^Z = \big((1 + X)(1 + Y)\big)^Z.$$
I believe that to be true, but I have no idea how to prove it.
Write
$$(1+X)^Z(1+Y)^Z= \sum_{i,j\ge0} f_{i,j}(Z)X^iY^j,$$
$$(1+X+Y+XY)^Z=\sum_{i,j\ge0}g_{i,j}(Z)X^iY^j.$$
Since $(1+X)^n(1+Y)^n=(1+X+Y+XY)^n$ for all naturals $n\ge0$, we may conclude that the coefficients $f_{ij}(n)=g_{ij}(n)$ agree on infinitely many $n$, hence $f_{i,j}(Z)=g_{i,j}(Z)$.