You can generate the Bernoullis through the combinatorics of permutahedra and graphical interpretations of surjections presented in OEIS-[A133314][1]  (cf. [A049019][2], [A019538][3], [A008292][4]) weighted by the reciprocal integers. (See also MOQ-[61252][5].)

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 from the Bernoullis to the reciprocal integers by the same weighted surjections.

But, with the o.g.f., using the normalized Bernoulli numbers, compositional inversion, rather than reciprocation enters the picture and, therefore, weighted noncrossing partitions and Dyck lattice paths (and myriad other related combinatoric structures). See the last paragraph of my answer to the MOQ referenced above.  

  [1]: https://oeis.org/A133314
  [2]: https://oeis.org/A049019
  [3]: https://oeis.org/A019538
  [4]: https://oeis.org/A008292
  [5]: http://mathoverflow.net/questions/61252/why-do-bernoulli-numbers-arise-everywhere