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 $$B(x) = e^{e^x-1}. $$
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....