How it can be proved that : $$\sum_{n=0}^{ ∞}B_{n}\left(x\right)\frac{t^{n}}{n!}=e^{x\left(e^{t}-1\right)}$$
Where $B_n$ is the $n^{th}$ complete Bell polynomial. I know that $$\sum_{n=k}^{∞ }S\left(n,k\right)\frac{t^{n}}{n!}=\frac{1}{k!}\left(e^{t}-1\right)^{k}$$
where $S\left(n,k\right)$ is Stirling number of the second kind,e.g. the number of way to partition a say with $n$ distinguished objects into $k$ nonempty partitions, but is it useful to prove my question?
The Bell polynomials are defined by \begin{eqnarray*} B_n(x)= \sum_{k=0}^{n} {n \brace k} x^k. \\ \end{eqnarray*}
Thus we have \begin{eqnarray*} \sum_{n=0}^{\infty} B_n(x) \frac{t^n}{n!} =\sum_{n=0}^{\infty} \sum_{k=0}^{n} {n \brace k} x^k \frac{t^n}{n!}. \\ \end{eqnarray*} Now inverting the order of the plums & using the formula stated in the question \begin{eqnarray*} \sum_{n=0}^{\infty} \sum_{k=0}^{n} {n \brace k} x^k \frac{t^n}{n!} &=& \sum_{k=0}^{\infty} \sum_{n=k}^{\infty} {n \brace k} \frac{t^n}{n!} x^k \\ &=& \sum_{k=0}^{\infty} \frac{(e^{t}-1)^k}{k!} x^k \\ &=& \color{red}{ \operatorname{exp}(x(e^t-1)) }. \end{eqnarray*}