Formula for Stirling numbers expressed with Bernoulli numbers?

So, nobody answer this, so I am going to extend what I wrote in the comments.

We know that $$\frac{(e^x-1)^k}{k!}=x^k\sum _{n\geq k}\sigma (n,k)\frac{x^{n-k}}{n!},$$ and that $$\frac{x}{e^x-1}=\sum _{n = 0}^{\infty}B_n\frac{x^n}{n!}.$$ So, to relate both we just manipulate the first equation to get $$\frac{1}{\left (1-(1-\frac{x}{e^x-1})\right )^k}=\sum _{n\geq k}\sigma (n,k)\frac{x^{n-k}k!}{n!},$$ expanding the LHS one gets $$\sum _{m=0}^{\infty}\binom{m+k-1}{k-1}\left (-\sum _{\ell = 1}^{\infty}\frac{B_{\ell}}{\ell !}x^{\ell}\right )^m.$$ Expanding that one gets at the end that $$\sigma (n,k)=\binom{n}{k}\sum _{m=0}^{\infty}\binom{m+k-1}{k-1}(-1)^m\sum _{x_1+\cdots +x_m=n-k}\binom{n-k}{x_1,\cdots ,x_m}\prod _{\ell = 1}^mB_{x_{\ell}}.$$ You can try a script to check this here