Why is the exponential of this formal power series well-defined? 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!
 A: 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
$.
