Determine Coefficient from Generating Functions Let $G\colon \mathbb{N}^3\to \mathbb{N}$ be such that 
$$
\sum_{n=1}^\infty \bigg(\sum_{i=1}^\infty x^i \bigg)^n \bigg(\sum_{j=1}^\infty y^j \bigg)^n\bigg( \sum_{k=1}^\infty z^k\bigg)^n = \sum_{i=1}^\infty \sum_{j=1}^\infty \sum_{k=1}^\infty G(i,j,k) x^i y^j z^k.
$$
How can I find a closed form for $G$?  This question is related to this one.  I am not well versed with generating functions.  Any hints are well appreciated.
 A: At first we derive a closed form of the generating function $\mathcal{G}(x,y,z)$ and then we extract the coefficient $G(i,j,k)$ from it.

Generating function: $\mathcal{G}(x,y,z)$
We obtain
  \begin{align*}
\mathcal{G}(x,y,z)&=\sum_{n=1}^\infty\left(\sum_{i=1}^\infty x^i \right)^n
\left(\sum_{j=1}^\infty y^j\right)^n\left(\sum_{k=1}^\infty z^k\right)^n\\
&=\sum_{n=1}^\infty\left(\frac{x}{1-x}\right)^n\left(\frac{y}{1-y}\right)^n\left(\frac{z}{1-z}\right)^n\tag{1}\\
&=\sum_{n=1}^\infty\left(\frac{xyz}{(1-x)(1-y)(1-z)}\right)^n\\
&=\frac{xyz}{(1-x)(1-y)(1-z)}\left(1-\frac{xyz}{(1-x)(1-y)(1-z)}\right)^{-1}\tag{2}\\
\end{align*}

We apply in (1) and (2) the formula for the geometric series expansion.

Coefficient extraction: $G(i,j,k)$
We denote with $[x^i]$ the coefficient of $x^i$ in a series. This way we can write
  \begin{align*}
G(i,j,k)&=[x^iy^jz^k]\mathcal{G}(x,y,z)\\
&=[x^i][y^j][z^k]\mathcal{G}(x,y,z)
\end{align*}
We obtain
  \begin{align*}
G(i,j,k)&=[x^iy^jz^k]\mathcal{G}(x,y,z)\\
&=\sum_{n=1}^\infty[x^i]\left(\frac{x}{1-x}\right)^n[y^j]\left(\frac{y}{1-y}\right)^n[z^k]\left(\frac{z}{1-z}\right)^n\tag{3}
\end{align*}
We have three similar parts and put the focus on the first one.
We obtain
  \begin{align*}
[x^i]\left(\frac{x}{1-x}\right)^n&=[x^i]x^n\sum_{p=0}^\infty\binom{-n}{p}(-x)^p\tag{4}\\
&=[x^{i-n}]\sum_{p=0}^\infty\binom{n+p-1}{p}x^p\tag{5}\\
&=\begin{cases}
\binom{i-1}{i-n}\qquad&\qquad 1\leq n\leq i\\
0\qquad&\qquad\text{otherwise}
\end{cases}
\end{align*}

Comment:


*

*In (4) we apply the binomial series expansion

*In (5) we use the rule $[x^{i-n}]A(x)=[x^i]x^nA(x)$. Note that $[x^{i-n}]$ is zero if $i<n$. We also use the binomial identity $\binom{-n}{p}=\binom{n+p-1}{p}(-1)^p$.

Putting all together in (3) we finally obtain
\begin{align*}
G(i,j,k)=\sum_{n=1}^{\min\{i,j,k\}}\binom{i-1}{i-n}\binom{j-1}{j-n}\binom{k-1}{k-n}\qquad\qquad 1\leq n\leq i,j,k
\end{align*}

