In the wikipedia article on the Faà di Bruno formula, one considers the composition of two "exponential/Taylor-like" series $\displaystyle f(x)=\sum_{n=1}^\infty {\frac{a_n}{n!}} x^n\; $ ($a_0=0$) and $\displaystyle g(x)=\sum_{n=0}^\infty {\frac{b_n}{n!}} x^n$
$$g\left(f(x)\right)=\sum_{n=0}^\infty{\frac{c_n}{n!}}x^n\qquad \text{where}\qquad c_n=\sum_{k=1}^n\sum_{\pi=\left\{B_1,\ldots,B_k\right\}} a_{\left|B_1\right|}\cdots a_{\left|B_k\right|} b_k \tag{1}\label{1}$$
and $\pi$ runs through the set of all partitions of the set $\{1, \cdots, n\}$ and $B_1, \cdots, B_k$ are the blocks of the partition $\pi$, and $\lvert B_j \rvert$ is the number of members of the $j$-th block, for $j= 1, \cdots , k$. $$ \vdots$$ $$ g(f(x)) = b_0+ \sum_{n=1}^\infty \frac{\sum_{k=1}^n b_k\, B_{n,k}(a_1,\ldots,a_{n-k+1})}{n!} x^n \tag{2}\label{2}$$
I am triple checking that $$ c_n =\sum_{k=1}^n\sum_{\pi=\left\{B_1,\ldots,B_k\right\}} a_{\left|B_1\right|}\cdots a_{\left|B_k\right|} b_k= \sum_{k=1}^n b_k\, B_{n,k}(a_1,\ldots,a_{n-k+1}) \tag{3}\label{3}$$ It seems to work in a first way 1. but not in a second way 2. ...
- As in the corresponding paragraph in wikipedia, let us start with the product of an "ordinary" series $\displaystyle F(x)=\sum_{p=0}^{+\infty} \alpha_p\, x^p$ $$\big(F(x)\big)^k= \sum_{n=0}^{+\infty} A_{k,n} x^n \qquad\text{where}\qquad A_{k,n}=\sum_{i_1+i_2 + \cdots + i_k=n} \alpha_{i_1}\,\alpha_{i_2}\,\cdots \, \alpha_{i_k}$$ (For $k=2$ one reckonizes the Cauchy product of series, and this generalization can be understood by saying that each $\alpha_{i_1}$ comes with the factor $x^{i_1}$ so that the condition $i_1+i_2 + \cdots + i_k=n$ on summation indices precisely select the terms in $x^n$) The trick is now to rewrite $A_{k,n}$ by regrouping the $\alpha_{i_i}, \alpha_{i_j}$ whenever $i_i=i_j$: $$A_{k,n}=\sum_{\genfrac{}{}{0pt}{1}{m_1+ m_2+ m_3+\cdots + m_n=k}{m_1+2\cdot m_2+3\cdot m_3+\cdots + n\cdot m_n=n}} C_{m_1,\cdots, m_n}\, \alpha_{1}^{m_1}\,\alpha_{2}^{m_2}\,\cdots \, \alpha_{n}^{m_n} $$ (Indeed $\alpha_{i_1}\,\alpha_{i_2}\,\cdots \, \alpha_{i_k}$ can always be written as $\alpha_{1}^{m_1}\,\alpha_{2}^{m_2}\,\cdots \, \alpha_{n}^{m_n} $ but the condition that there was $k$ factors has now to be imposed by the condition $m_1+ m_2+ m_3+\cdots + m_n=k$. Moreover, the condition that $i_1+i_2 + \cdots + i_k=n$ now becomes $m_1+2\cdot m_2+3\cdot m_3+\cdots + n\cdot m_n=n$ as there are $m_i$ factor $\alpha_i$.) Finally, for a given $(m_1,\cdots, m_n),\ C_{m_1,\cdots, m_n}$ counts the number of occurence of $\alpha_{1}^{m_1}\,\alpha_{2}^{m_2}\,\cdots \, \alpha_{n}^{m_n}$ in the original expression of $A_{k,n}$. Let $(m_1,\cdots, m_n)$ satisfy the summation conditions and consider $i_1=i_2=\cdots=i_{m_1}=1,\; i_{m_1+1}=i_{m_1+2}=\cdots=i_{m_1+m_2}=2$ etc. We see that the number of ordered lists $(i_1,\cdots, i_k)$ corresponding to that specific $(m_1,\cdots, m_n)$ is given by the multinomial coefficient $$ C_{m_1,\cdots, m_n} = \frac{k!}{m_1!\, m_2!\, \cdots\, m_n!} \tag{4}\label{4}$$ so $$ A_{k,n}=\sum_{\genfrac{}{}{0pt}{1}{m_1+ m_2+ m_3+\cdots + m_n=k}{m_1+2\cdot m_2+3\cdot m_3+\cdots + n\cdot m_n=n}} \frac{k!}{m_1!\, m_2!\, \cdots\, m_n!}\, \alpha_{1}^{m_1}\,\alpha_{2}^{m_2}\,\cdots \, \alpha_{n}^{m_n} = \frac{k!}{n!}\, B_{n,k} \big( \alpha_{1} , 2!\, \alpha_{2} ,\, \cdots \, , (n-k+1)!\, \alpha_{n-k+1}\big)$$ if one compares with the definition of partial Bell polynomial, cf. also ordinary Bell polynomial. Finally (the hypothesis $a_0=0$ on $f$ is such that $\big(F(x)\big)^k$ has valuation $k$, so there is no $x^n$ term for $k>n$), assuming that one has absolute convergence so that one can exchange summation $$g(f(x)) = \sum_{k=0}^\infty {\frac{b_k}{k!}} \big(F(x)\big)^k = b_0 + \sum_{k=1}^\infty \frac{b_k}{k!} \sum_{n=k}^{+\infty} A_{k,n}\, x^n = b_0 + \sum_{n=1}^\infty \sum_{k=1}^{n}\frac{b_k}{n!}\, B_{n,k} \big( \alpha_{1} , 2!\, \alpha_{2} ,\, \cdots \, , (n-k+1)!\, \alpha_{n-k+1}\big)\, x^n $$ which is (\ref{2}) if $\displaystyle f(x)=\sum_{p=0}^{+\infty} \alpha_p\, x^p = \sum_{p=0}^{+\infty} \frac{a_p}{p!}\, x^p$, i.e. $\ p!\cdot \alpha_p = a_p$.
- Now directly for $\displaystyle f(x)=\sum_{n=1}^\infty {\frac{a_n}{n!}} x^n$, one has $$\big(f(x)\big)^k= \sum_{n=0}^{+\infty} \frac{\tilde{A}_{k,n}}{n!} x^n\qquad\text{with}\qquad \tilde{A}_{k,n}=\sum_{i_1+i_2 + \cdots + i_k=n} \frac{n!}{i_1!\, i_2!\, \cdots\, i_k!} a_{i_1}\,a_{i_2}\,\cdots \, a_{i_k} $$ The trick now is to make the multinomial coefficient appear by summing over the set of partitions of $\left\lbrace 1, 2, \cdots , n\right\rbrace $ in $k$ (non-empty) blocks: $$\sum_{i_1+i_2 + \cdots + i_k=n} \frac{n!}{i_1!\, i_2!\, \cdots\, i_k!} a_{i_1}\,a_{i_2}\,\cdots \, a_{i_k} = \sum_{\pi=\left\{B_1,\ldots,B_k\right\}} a_{\left|B_1\right|}\cdots a_{\left|B_k\right|} \label{5}\tag{5}$$ For a given $k$-tuple $(i_1,\cdots , i_k)$, let $B_1 \cup \cdots \cup B_k= \{1, \cdots , n\}$ an adapted partitions: the $i_1! \,\cdots \, i_k!$ permutations leaving each block $B_j$ unchanged form the stabilizer subgroup, so that there are $\frac{n!}{i_1!\, i_2!\, \cdots\, i_k!}$ partitions such that $|B_{j}|=i_j$. But then $$\tilde{A}_{k,n}=\sum_{\pi=\left\{B_1,\ldots,B_k\right\}} a_{\left|B_1\right|}\cdots a_{\left|B_k\right|} = B_{n,k}\big(a_{1}\,a_{2},\,\cdots \, , a_{n-k+1} \big) \label{6}\tag{6}$$ Finally $$g(f(x)) = \sum_{k=0}^\infty {\frac{b_k}{k!}} \big(f(x)\big)^k = b_0 + \sum_{k=1}^\infty \frac{b_k}{k!} \sum_{n=k}^{+\infty} \frac{\tilde{A}_{k,n}}{n!}\, x^n = b_0 + \sum_{n=1}^\infty \sum_{k=1}^{n}\frac{b_k}{k!}\, \frac{B_{n,k} \big(a_{1}\,a_{2},\,\cdots \, , a_{n-k+1} \big)}{n!}\, x^n \label{Err1}\tag{Err1}$$ Hence, we find a different result, namely $$c_n = \sum_{k=1}^{n}\frac{b_k}{k!}\, B_{n,k} \big(a_{1}\,a_{2},\,\cdots \, , a_{n-k+1} \big) \label{Err2}\tag{Err2}$$
It seems that the mistake is at equation (6) as it seems that we do pay attention to the order of the label $(B_1, B_2,\cdots , B_k)$ (at least, it is the case in l.h.s. of (5), so also for its r.h.s.). So (6) has to be replaced by $$\tilde{A}_{k,n}=\sum_{\pi=\left\{B_1,\ldots,B_k\right\}} a_{\left|B_1\right|}\cdots a_{\left|B_k\right|} = k!\, B_{n,k}\big(a_{1}\,a_{2},\,\cdots \, , a_{n-k+1} \big) \label{6b}\tag{6b}$$ We do find $$g(f(x)) = \sum_{k=0}^\infty {\frac{b_k}{k!}} \big(f(x)\big)^k = b_0 + \sum_{k=1}^\infty \frac{b_k}{k!} \sum_{n=k}^{+\infty} \frac{\tilde{A}_{k,n}}{n!}\, x^n = b_0 + \sum_{n=1}^\infty \frac{c_n}{n!}\, x^n $$ with $$c_n = \sum_{k=1}^{n} b_k\, B_{n,k} \big(a_{1}\,a_{2},\,\cdots \, , a_{n-k+1} \big) \label{Corr}\tag{Corr}$$