Product of three generating functions I have proven that if $f$ and $g$ are two generating functions then
$\displaystyle fg=\left\{\sum_r\binom{n}{r} a_rb_{n-r}\right\}_{n=0}^{\infty}$
In generatingfunctionology page 45 it is mentioned that
$\displaystyle fgh=\left\{ \sum_{r+s+t=n} \frac{n!}{r!s!t!} a_rb_sc_t \right\}_{n=0}^{\infty}$
I am confused by this triple product, since the book does not give a "formal proof" I would like to know how to generate this formula since the multiple product is directly tested from this
 A: What have you proven seems not right, as $fg$ is a power series and $\left\{\sum_r\binom{n}{r} a_rb_{n-r}\right\}_{n=0}^{\infty}$ is a not well-defined sequence, maybe you mean that the coefficients of the exponential power series $fg$ are in 1-to-1 correspondence with the sequence  $\left\{\sum_{r=0}^n\binom{n}{r} a_rb_{n-r}\right\}_{n=0}^{\infty}$.
Indeed we have that
$$
f(x):=\sum_{k\geqslant 0}a_k \frac{x^k}{k!},\quad g(x):=\sum_{k\geqslant 0}b_k\frac{x^k}{k!}\\
\therefore\quad  f(x)\cdot g(x)=\sum_{n\geqslant 0}c_n \frac{x^n}{n!},\quad \text{ where }\quad c_n:=\sum_{k=0}^n \binom{n}{k}a_k b_{n-k}
$$
formally. Then for $h(x):=\sum_{n\geqslant 0}d_n \frac{x^n}{n!}$ and $\ell (x):=f(x)\cdot g(x)$ applying the previous result we get
$$
\ell (x)\cdot h(x)=\sum_{r\geqslant 0}e_r\frac{x^r}{r!},\quad e_r:=\sum_{n=0}^r \binom{r}{n}c_nd_{r-n}
$$
But
$$
\sum_{n=0}^r \binom{r}{n}c_nd_{r-n}=\sum_{n=0}^r\sum_{k=0}^n \binom{r}{n} \binom{n}{k}a_k b_{n-k}d_{r-n}\\
=\sum_{0\leqslant n\leqslant r}\frac{r!}{(r-n)!(n-k)!k!}a_k b_{n-k}d_{r-n}\\
=\sum_{j,k,l\in \mathbb{N}\cup \{0\}: j+k+l=r}\frac{r!}{j!k!l!}a_jb_kd_l
$$
where the last result follows from noticing that, from the condition $0\leqslant n\leqslant r$ then the coefficients $r-n, n-k,k$ range exactly between all distinct possible sums of non-negative integers that add up to $r$.

ADDITION: a cleanest and more complete proof can be given using induction, however the notation can be slightly cumbersome. Let the induction hypothesis
$$
\text{ If }\, f_k(x)=\sum_{m\geqslant 0}c_{m,k}\frac{x^m}{m!}\, \text{ for each }\, k\in\{1,\ldots,n\}\, \text{ then }\, 
\prod_{k=1}^n f_k(x)=\sum_{m\geqslant 0}p_m \frac{x^m}{m!}\quad \\
\text{ where }\quad  p_m=\sum_{\substack{j_1+j_2+\ldots +j_n=m\\j_1,j_2,\ldots ,j_n\in \mathbb N\cup\{0\}}} \frac{m!}{j_1!\cdot j_2!\cdots j_n!}\prod_{k=1}^n c_{j_k,k}
$$
Then setting $g(x):=\prod_{k=1}^n f_k(x)$ and with base case $n=2$ we get that
$$
\begin{align*}
g(x)\cdot f_{n+1}(x)&=\sum_{m\geqslant 0}p'_m \frac{x^m}{m!}\quad  \text{  where }\\ p'_m&=\sum_{r+j_{n+1}=m} \frac{m!}{r!\cdot j_{n+1}!}p_r\cdot c_{j_{n+1},n+1}\\
&=\sum_{r+j_{n+1}=m} \frac{m!}{r!\cdot j_{n+1}!}\left(\sum_{\substack{j_1+j_2+\ldots +j_n=r\\j_1,j_2,\ldots ,j_n\in \mathbb N\cup\{0\}}} \frac{r!}{j_1!\cdot j_2!\cdots j_n!}\prod_{k=1}^n c_{j_k,k}\right)\cdot c_{j_{n+1},n+1}\\
&=\sum_{\substack{j_1+j_2+\ldots +j_{n+1}=m\\j_1,j_2,\ldots ,j_{n+1}\in \mathbb N\cup\{0\}}} \frac{m!}{j_1!\cdot j_2!\cdots j_{n+1}!}\prod_{k=1}^{n+1} c_{j_k,k}
\end{align*}
$$
∎
