The generalized product rule complicates putting series coefficients into closed or hypergeometric form. There are 2 forms with Lagrange $n$th derivative notation and the multinomial $\binom n{n_1,\dots,n_j}$; the $\displaystyle\sum_{\sum\limits_jn_j=n}$ version:
$$\left(\prod_{k=1}^m f_k\right)^{(n)}=\sum_{\sum\limits_jn_j=n} \binom n{n_1,\dots,n_m} \prod_{k=1}^m f_k^{(n_k)}$$
and the the $\displaystyle\sum_{n_1,\dots,n_j=0}^n$ version with Kronecker delta $\delta_{a,b}$:
$$\left(\prod_{k=1}^m f_k\right)^{(n)}=\sum_{n_1=0}^n\dots\sum_{n_m=0}^n\delta_{n,\sum\limits_{j=1}^mn_j} \binom n{n_1,\dots,n_m} \prod_{k=1}^m f_k^{(n_k)}.$$
But evaluating these sum by sum is complicated for given $f_k(x)$, so here is an alternate multinomial theorem without a single sum over a multivariable restriction nor a tensor function via repeated generalized Leibniz rule:
$$(fg)^{(n)}=\sum_{k=0}^n\binom nkf^{(k)}g^{(n-k)}$$
Trying with $5$ functions, one gets: $$\begin{align}(f_1f_2f_3f_4f_5)^{(n)}=\sum_{n_1=0}^n\binom n{n_1}f_1^{(n-n_1)}\sum_{n_2=0}^{n_1}\binom{n_1}{n_2}f_2^{(n_1-n_2)}(f_3f_4f_5)^{(n_2)}= \sum_{n_1=0}^n\binom n{n_1}f_1^{(n-n_1)}\sum_{n_2=0}^{n_1}\binom{n_1}{n_2}f_2^{(n_1-n_2)}\sum_{n_3=0}^{n_2}\binom{n_2}{n_3}f_3^{(n_2-n_3)}\sum_{n_4=0}^{n_3}\binom{n_3}{n_4}f_4^{(n_3-n_4)}f_5^{(n_4)}=\sum_{n_1=0}^n\sum_{n_2=0}^{n_1}\sum_{n_3=0}^{n_2}\sum_{n_4=0}^{n_3}\frac{n!f_1^{(n-n_1)} f_2^{(n_1-n_2)} f_3^{(n_2-n_3)} f_4^{(n_3-n_4)}f_5^{(n_4)}}{(n-n_1)!(n_1-n_2)!(n_2-n_3)!(n_3-n_4)!n_4!}\end{align}$$
Therefore:
$$\boxed{\left(\prod_{k=1}^mf_k\right)^{(n)}\mathop=^?\sum_{n_1=0}^n\sum_{n_2=0}^{n_1}\dots\sum_{n_{m-1}=0}^{n_{m-2}}\binom n{n_1}f_1^{(n-n_1)}f_m^{(n_{m-1})}\prod_{k=2}^{m-1}\binom{n_{k-1}}{n_k}f_k^{(n_{k-1}-n_k)}=\sum_{n_1=0}^n\sum_{n_2=0}^{n_1}\dots\sum_{n_{m-1}=0}^{n_{m-2}}\frac{n!f_1^{(n-n_1)}f_m^{(n_{m-1})}}{(n-n_1)!n_{m-1}!}\prod_{k=2}^{m-1}\frac{f_k^{(n_{k-1}-n_k)}}{(n_{k-1}-n_k)!}}\tag1$$
Similarly, but without $(n_{k-1}-n_k)$ in any but $1$ $n$th derivative and factorial:
$$(f_1f_2f_3f_4f_5)^{(n)}=\sum_{n_1=0}^n\binom n{n_1}f_1^{(n_1)}(f_2f_3f_4f_5)^{(n-n_1)}= \sum_{n_1=0}^n\binom n{n_1}f_1^{(n_1)}\sum_{n_2=0}^{n-n_1}\binom{n-n_1}{n_2}f_2^{(n_2)}\sum_{n_3=0}^{n-n_1-n_2}\binom{n-n_1-n_2}{n_3}f_3^{(n_3)}\sum_{n_4=0}^{n-n_1-n_2-n_3}\binom{n-n_1-n_2-n_3}{n_4} f_4^{(4)} f_5^{(n-n_1-n_2-n_3-n_4)}=\sum_{n_1=0}^n\sum_{n_2=0}^{n-n_1}\sum_{n_3=0}^{n-n_1-n_2}\sum_{n_4=0}^{n-n_1-n_2-n_3}\frac{n!f_1^{(n_1)}f_2^{(n_2)}f_3^{(n_3)}f_4^{(n_4)}f_5^{(n-n_1-n_2-n_3-n_4)}}{n_1!n_2!n_3!n_4!(n-n_1-n_2-n_3-n_4)!}$$
Therefore: $$\boxed{\left(\prod_{k=1}^mf_k\right)^{(n)}\mathop=^?\sum_{n_1=0}^n\sum_{n_2=0}^{n-n_1}\dots\sum_{n_{m-1}=0}^{n-\sum\limits_{k=1}^{m-2}n_k}\frac{n!f_m^{\left(n-\sum\limits_{k=1}^{m-1}n_k\right)}}{\left(n-\sum\limits_{k=1}^{m-1}n_k\right)!}\prod_{k=1}^{m-1}\frac{f_k^{(n_k)}}{n_k!}= \sum_{n_1=0}^n\sum_{n_2=0}^{n-n_1}\dots\sum_{n_{m-1}=0}^{n-\sum\limits_{k=1}^{m-2}n_k}\binom n{n_1,\dots,n_{m-1}}\frac{f_m^{\left(n-\sum\limits_{k=1}^{m-1}n_k\right)}}{\left(n-\sum\limits_{k=1}^{m-1}n_k\right)!}\prod_{k=1}^{m-1}f_k^{(n_k)}}\tag2$$
Are $(1),(2)$ correct and fully simplified?
Best Answer
Both expressions (1) and (2) look fine. Expression (1) could be slightly simplified, if we write $n_0$ instead of $n$. This way (1) becomes \begin{align*} \color{blue}{\left(\prod_{k=1}^n f_k\right)^{\left(n_0\right)}} &=n_0!\sum_{n_1=0}^{n_0}\sum_{n_2=0}^{n_1}\cdots\sum_{n_{m-1}=0}^{n_{m-2}} \frac{f_{m}^{\left(n_{m-1}\right)}}{n_{m-1}!} \prod_{k=1}^{m-1}\frac{f_n^{\left(n_{k-1}-n_k\right)}}{\left(n_{k-1}-n_{k}\right)!}\\ &\color{blue}{=n_0!\sum_{0\leq n_{m-1}\leq n_{m-2}\leq \cdots\leq n_1\leq n_{0}} \frac{f_{m}^{\left(n_{m-1}\right)}}{n_{m-1}!} \prod_{k=1}^{m-1}\frac{f_n^{\left(n_{k-1}-n_k\right)}}{\left(n_{k-1}-n_{k}\right)!}}\\ \end{align*}