You can generate the Bernoullis through the combinatorics of permutahedra and graphical interpretations of surjections presented in OEIS-A133314 (cf. A049019, A019538, A008292) weighted by the reciprocal integers. (See also MOQ-61252.)
This is equivalent to determining the reciprocal of the exponential generating function
$$\frac{e^t-1}{t}=1+\frac{1}{2}t+ \frac{1}{3}\frac{t^2}{2!}+ \frac{1}{4}\frac{t^3}{3!}+\cdots\;\;.$$
Naturally, it's an involution, so you can go in the reverse direction, which seems to be what you have effectively done.