Prove that $$\sum_{k=0}^n s(n,k)*B(k) = 1$$ where s(n,k) - Stirling number of the first kind; B(k) - Bell number.
I've tried to use $$B(k)=\sum_{j=0}^kS(k,j)$$ where S(n,j) - Stirling numbers of the second kind; However, the proof of $$\sum_{k=0}^n \sum_{j=0}^ks(n,k)*S(k,j) = 1$$ isn't easier and i stuck dealing with induction step for n+1.
I'm allowed to use only simplest facts such as stirlings reccurence and Bell's definition + those that are easy-to-prove.
Using formal power series and standard EGFs we get for the sum
$$\sum_{k=0}^n (-1)^{n+k} n! [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k k! [w^k] \exp(\exp(w)-1) \\ = (-1)^n n! [z^n] \sum_{k=0}^n (-1)^{k} \left(\log\frac{1}{1-z}\right)^k [w^k] \exp(\exp(w)-1).$$
Now $$\left(\log\frac{1}{1-z}\right)^k = z^k + \cdots$$ and hence we may extend $k$ beyond $n$ since these values do not contribute to the coefficient on $[z^n]$, obtaining
$$(-1)^n n! [z^n] \sum_{k\ge 0} (-1)^{k} \left(\log\frac{1}{1-z}\right)^k [w^k] \exp(\exp(w)-1) \\ = (-1)^n n! [z^n] \exp\left(\exp\left(-\log\frac{1}{1-z}\right)-1\right) \\ = (-1)^n n! [z^n] \exp((1-z)-1) = (-1)^n n! [z^n] \exp(-z) = (-1)^n (-1)^n \\ = 1.$$