TITLE: Computing the exponential generating function of the Bell numbers.
QUESTION [5 upvotes]: I am trying to compute the exponential generating function of the Bell numbers $B_{n+1} = \sum_{k=0}^n \binom nkB_k, B_1=1$. So far I have
\begin{align}
B(x) &= \sum_{n=0}^\infty B_n\frac{x^n}{n!}\\
&= \sum_{n=0}^\infty \sum_{k=0}^n \binom nk B_k \frac{x^n}{n!}\\
&=\sum_{k=0}^\infty B_k\sum_{n=k}^\infty \binom nk \frac{x^n}{n!},
\end{align}
where we can interchange the order of summation by monotone convergence or Tonelli's theorem. But I have no idea how to compute $\sum_{n=k}^\infty \frac{x^n}{k!(n-k)!}$. According to Mathematica,
$$
\sum_{n=k}^\infty \frac{x^n}{k!(n-k)!}  = \frac{e^x x^k}{k!}.
$$
As shown in the answer,
$$
\sum_{n=k}^\infty \frac{x^n}{k!(n-k)!} = \sum_{n=0}^\infty \frac{x^{n+k}}{k!n!} = x^k \sum_{n=0}^\infty \frac{x^n}{n!} = \frac{x^ke^x}{k!}.$$
Hence
$$
B(x) = \sum_{k=0}^\infty B_k \frac{e^xx^k}{k!} = e^x \sum_{k=0}^\infty B_k\frac{x^k}{k!} = e^x B(x).
$$
But this does not make sense, as $B(x) = e^xB(x)$ implies that $B(x)=0$. What error have I made?
Edit: It turns out that $$B(x) = 1 + \sum_{n=0}^\infty\sum_{k=0}^\infty \binom nk B_k\frac{x^n}{n!},$$ so what I computed above was actually $B'(x)$. This yields the differential equation $B'(x) = e^x B(x)$, from which $B(x) = Ce^{e^x}$. The condition $B(0)=1$ yields $C=\frac1e$, so that $$B(x) = e^{e^x-1}. $$

REPLY [5 votes]: There's an error in your second line. Using $B_0=1$, you should get
$$B(x)=1+\sum_{n=0}^\infty\sum_{k=0}^n\binom nkB_k\frac{x^{n+1}}{(n+1)!}.$$
I would now differentiate....