You can also approach a proof of a more general relation using the generalized Dobinski formula:

$$f(\phi.(x))= e^{-x}exp(a.x)=exp(-(1-a.)x),$$

where $(\phi.(x))^n=\phi_n(x)$ is the $n$th Bell polynomial with $B_n=\phi_n(1)$ and $(a.)^n=a_n=f(n).$

Then $$\sum_{k=0}^\infty\phi_k(x) t^k=\frac{1}{1-\phi.(x)t}=e^{-x}\sum_{n=0}^\infty\frac{1}{1-nt}\frac{x^n}{n!}$$
$$=\sum_{n=0}^\infty \frac{x^n}{n!}\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}\frac{1}{1-jt}.$$

And, the last finite difference expression is the partial fraction expansion of  $n!\prod_{j=1}^n \frac{t}{1-jt}$, so

 $$\sum_{k=0}^\infty\phi_k(x) t^k=1+\sum_{n=1}^\infty x^n \prod_{j=1}^n \frac{t}{1-jt},$$

which reduces to the illustrated formula when $x=1$.

Other proofs, including those alluded to in other answers, can be found in W. Lang's [notes][1]. 

The generalized Dobinski relation is a consequence of
 $$f(\phi.(:xD:))x^n=f(xD)x^n=f(n)x^n=a_n x^n=(a.x)^n,$$
where $D=d/dx$ and $(:xD:)^k=x^kD^k$ by definition, so

 $$f(\phi.(:xD:))e^x=e^xf(\phi.(x))=f(xD)e^x=e^{a.x}.$$


  [1]: http://oeis.org/A071919/a071919.pdf