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.

  [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