I don't understand how the definition for the generating function of Bell polynomials, i.e. $\Phi(t):=\exp(\sum_{k=1}^\infty x_k \frac{t^k}{k!})$, makes sense. If we write $F:=\sum_{k=1}^\infty x_k \frac{t^k}{k!}$, then $F\in \mathbb{C}[X_1,X_2,...][[t]]$ and $\Phi=\exp(F)$ but I don't get how one defines $\exp(F)$.
Let $\mathbb{C}[X_1,X_2,...][[t]]$ be equipped with the discrete topology. The above definition makes sense only if the sequence $\{\sum_{k=1}^n \frac{F^k}{k!}\}$ is convergent, but it is very unclear to me.
Or think of it in this way.
Define $\phi(u):=\sum_{n=0}^\infty \frac{(Fu)^n}{n!}$ so that $\phi \in \mathbb{C}[X_1,X_2,...][[t]][[u]]$.
$\Phi$ is well- define iff the evaluation of $\phi$ at $1$, i.e. $\phi(1)$ should be well-defined. But how?
Thank you in advance!
Here is a method (not original by me) to find $p(t) =\exp(f(t)) =\exp(\sum_{k=1}^\infty a_k \frac{t^k}{k!}) $.
If $p(t) =\exp(f(t)) $, differentiating, $p'(t) =f'(t)\exp(f(t)) =f'(t)p(t) $.
If $p(t) =\sum_{i=1}^{\infty} p_i \frac{ t^i}{i!} $ (we want to find the $p_i$), then $p'(t) =\sum_{i=1}^{\infty} ip_i \frac{t^{i-1}}{i!} =\sum_{i=0}^{\infty} p_{i+1} \frac{t^{i}}{i!} $ and $f'(t) =\sum_{k=1}^\infty ka_k \frac{t^{k-1}}{k!} =\sum_{k=0}^\infty a_{k+1} \frac{t^k}{k!} $.
Multiplying the power series,
$\begin{array}\\ p'(t) &=\sum_{n=0}^{\infty} p_{n+1} \frac{t^{n}}{n!}\\ \text{and}\\ f'(t)p(t) &=\sum_{k=0}^\infty a_{k+1} \frac{t^k}{k!}\sum_{i=0}^{\infty} p_i \frac{ t^i}{i!} \qquad\text{(with } p_0 = 0)\\ &=\sum_{k=0}^\infty\sum_{i=0}^{\infty} a_{k+1}p_i \frac{t^{k+i}}{k!i!}\\ &=\sum_{n=0}^\infty\sum_{k+i=n} a_{k+1}p_i \frac{t^{n}}{k!i!}\\ &=\sum_{n=0}^\infty\sum_{i=0}^n a_{n-i+1}p_i \frac{t^{n}}{(n-i)!i!} \qquad(k = n-i)\\ &=\sum_{n=0}^\infty\frac{t^n}{n!}\sum_{i=0}^n a_{n-i+1}p_i \frac{n!}{(n-i)!i!}\\ &=\sum_{n=0}^\infty\frac{t^n}{n!}\sum_{i=0}^n a_{n-i+1}p_i \binom{n}{i}\\ \end{array} $
Equating coefficients of $t^n$, $p_{n+1} =\sum_{i=0}^n a_{n-i+1}p_i \binom{n}{i} =\sum_{i=1}^n a_{n-i+1}p_i \binom{n}{i} $ since $p_0 = 0$.
To get $p_1$, set $t=0$ to get $p_1 =p'(0) =f'(0)p(0) =a_1\exp(f(0)) =a_1 $.