Prove that $\sum_{k=1}^{n} \frac{(-1)^{k}}{k}B(n,k) = 0 $ for all integer $n\geq 2$ For all $ n, k\geq 1$,  define $B(n,k) =  \sum_{a_{1}+a_{2} +\dots+a_{k} =n} \frac{1}{a_{1}!a_{2}! \cdots a_{k}! }$, where $ a_{1},a_{2},\dots,a_{k} $ are positive integers.
For example: $B(3,2) =  \frac{1}{1!2!} + \frac{1}{2!1!} =1$ because we can write $3 $ as  $2+1$ and $1+2$.
Another example:
 $$B(5,3) =  \frac{1}{1!1!3!} + \frac{1}{1!3!1!} + \frac{1}{3!1!1!}+\frac{1}{1!2!2!}+ \frac{1}{2!1!2!}+ \frac{1}{ 2!2!1!} =\frac{5}{4}. $$ 

Prove that  $\sum_{k=1}^{n} \frac{(-1)^{k}}{k}B(n,k) = 0 $  for all  integer $n\geq 2$.

I tried to approach this question by counting numbers of functions (onto), but someone said they did it by observing the coefficients of  $f(x) = \ln(x+1) ,g(x) = e^{x}-1$. I don't really understand how this is helpful. Can anyone provide relevant pointers?

It is well-known that
$$
e^x = \sum_{a=0}^\infty \frac{x^a}{a!}.
$$
This implies that
$$
(e^x-1)^k = \sum_{n=0}^\infty B(n,k) x^n.
$$
It is also known that
$$
\ln (1+x) = -\sum_{k=1}^\infty (-1)^k \frac{x^k}{k}.
$$
Therefore
$$
\ln (1 + (e^x-1)) = -\sum_{k=1}^\infty (-1)^k \frac{(e^x-1)^k}{k} = -\sum_{k=1}^\infty \sum_{n=0}^\infty (-1)^k \frac{B(n,k) x^n}{k}=-\sum_{n=0}^\infty x^n \sum_{k=1}^\infty \frac{(-1)^k}{k} B(n,k).
$$
On the other hand,
$$
\ln(1 + (e^x-1)) = \ln(e^x) = x.
$$
The proof now follows by comparing coefficients.
