So I have the function $ g(x) = \sum_{k=0}^{\infty} \frac{B_k}{k!}x^k $ where $B_k$ is the $k$'th Bernouli number.
I want to show that $g(x)(e^{x} - 1) = x$
I tried showing this by developing the taylor expansion for $e^x$ and by taking the first term from it and by performing index manipulation.
I got that the taylor expansion for $e^x$ after substructing $1$ is $\sum_{k=0}^{\infty}\frac{x^{k+1}}{(k+1)!}$
However, I wasnt able to show the equation above.