Is there a formula which expresses any of the Stirling numbers (1st or 2nd kind) in terms of the Bernoulli numbers? For example, here is the reverse
$$ B_k=\sum_{m=0}^{k} (-1)^m \frac{m!}{m+1}\sigma(k,m) $$
where $\sigma(k,m)$ is the Stirling number of the 2nd kind
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