In the OEIS entry for Bell numbers, there appears a generating function
$$\sum_{k=0}^\infty B_k t^k = \sum_{r=0}^\infty \prod_{i=1}^r \frac{t}{1-it}$$
However, I could not locate any proof of reference for this formula. The contributor informs me that he discovered it by experimentation.
I would appreciate any further information on this generating function.
The proof is given, for example, in http://www.sciencedirect.com/science/article/pii/S0097316503000141 (Bell numbers, their relatives, and algebraic differential equations, by Martin Klazar). Namely it is proved that the generating function $B(t)=\sum\limits_{n=0}^\infty B_nt^n$ satisfies the functional equation $$B(t)=1+\frac{t}{1-t}B\left(\frac{t}{1-t}\right).$$ Iterating this equation, we get (Klazar calls it the classical expansion of B(t)) $$B(t)=\sum\limits_{n=0}^\infty \frac{t^n}{(1-t)(1-2t)\cdots(1-nt)}.$$
Let $S_{n, k}$ denote the number of partitions of a set of size $n$ into $k$ partitions (the Stirling numbers of the second kind), so that $B_n = \displaystyle \sum_{k=0}^n S_{n, k}$ and hence, exchanging the order of summation, we have
$$\sum_{n=0}^{\infty} B_n t^n = \sum_{n=0}^{\infty} \sum_{k=0}^n S_{n, k} t^n = \sum_{k=0}^{\infty} t^k \sum_{n=k}^{\infty} S_{n, k} t^{n-k}.$$
A standard identity whose proof I used to know but have now forgotten asserts that
$$\sum_{n=k}^{\infty} S_{n, k} t^{n-k} = \frac{1}{(1 - t)(1 - 2t)\cdots(1 - kt)}$$
and this gives your formula. This is identity 1.94c in the second edition of Stanley's Enumerative Combinatorics, Volume I.
As a general remark, we may see the "Exponential to Ordinary" transformation of generating functions, $$ f(x):=\sum_{r=0}^\infty a_rx^r/r!\mapsto \tilde f(t):=\sum_{r=0}^\infty a_rt^r, $$ as an operator $\mathbb{C}[[x]]\to \mathbb{C}[[t]]$ . Since $r!t^r=\int_0^\infty (tx)^r e^{-x}dx$, the transformation can be computed analytically as $$\tilde f(t)=\int_0^\infty f(tx)e^{-x}dx,$$ at least for suitably convergent $f(x)$, and for special values of $t$. If the RHS is a convergent series $\sum_{r=0}^\infty b_rt^r$, and the equality holds, for a set of values $t$ which accumulates within the disk of convergence, the identity of series $ \sum_{r=0}^\infty a_rt^r=\sum_{r=0}^\infty b_rt^r$ is then established by the principle of isolated zeros.
For instance, we may compute the transform of $f_r(x):=(e^x-1)^r$ for real negative values of $t$ in terms of the Euler's Beta function by a change of variable in the integral:
$$\tilde f_r(t)=\int_0^\infty (e^{tx}-1)^re^{-x}dx=(-1)^{r+1}t^{-1}\int_0^1(1-u)^r u^{-1/t-1}du=$$
$$=(-1)^{r+1}t^{-1}\frac{\Gamma(r+1)\Gamma(-1/t)}{\Gamma(r+1-1/t)}=\frac{r!t^r}{(1-t)\dots(1-rt)}\ .$$
This computation gives your identity, since for the egf $f(x)$ of the $B_r$'s we have $f(x)=e^{e^x-1}=\sum_{r=0}^\infty\frac{1}{r!}f_r(x)$ (in the sense of formal power series) so that $\tilde f(t)=\sum_{r=0}^\infty \frac{t^r}{(1-t)\dots(1-rt)}\ .$
Incidentally, note that by an analogous computation you may, more generally, compute the ordinary gf of the Stirling polynomials of the second kind, $B_r(z) :=\sum_{r=0}^nS(n,r)z^r$, starting by their egf $e^{z(e^x-1)}$.
rmk. I think a more natural proof is, computing directly the ogf of the Stirling polynomials $B_r(z):=\sum_{r=0}^nS(n,r)z^r$, which consists just in translating the inductive relation $S(n+1,k)=kS(n,k)+S(n,k−1)$ in the language of formal power series, and then putting $z=1$, as done in other answers. The present answer is meant as an example of a general alternative approach.
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 $s\geq 0$):
$$\sum_{k=0}^\infty\phi_k((s)_.) t^k=\sum_{k=0}^\infty s^k t^k=\frac{1}{1-st}$$ $$=\sum_{n=0}^\infty(-1)^n \binom{s}{n}\sum_{j=0}^n(-1)^j\binom{n}{j}\frac{1}{1-jt}$$ $$=1+\sum_{n=1}^\infty (s)_n \prod_{j=1}^n \frac{t}{1-jt}.$$
