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 $(a.)^n=a_n=f(n)=f(x)|_{x=n}.$

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.

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

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

The umbral compositional inverse of the Bell / Touchard polynomial $\phi_n(x)$ is the falling factorial / Pochhammer symbol $(s)_n=s!/(s-n)!$, i.e., $\phi_n((s).)=s^n$ and $(\phi(x).)_n=x^n$, so a dual equation shadows that of the ordinary generating function above (for $t \leq0$ and $s\geq 0$):

$$\sum_{k=0}^\infty\phi_k((s)_.) t^k=\sum_{k=0}^\infty s^k t^k=\frac{1}{1-st}$$
$$=\sum_{n=0}^\infty(-1)^n \binom{s}{n}\sum_{j=0}^n(-1)^j\binom{n}{j}\frac{1}{1-jt}$$
$$=1+\sum_{n=1}^\infty (s)_n \prod_{j=1}^n \frac{t}{1-jt}.$$