I am having trouble showing the following
$$B_n^{(-m)}=\sum_{j=1}^n{\beta_n^{(j)}m^j}; n>0$$
where $B_n^{(k)}$ are generalized Bernoulli numbers and are generated by
$$\left(\frac{x}{e^x-1}\right)^k=\sum_{n=0}^\infty B_n^{(k)}\frac{x^n}{n!}$$
and
$$\beta_n^{(j)}=\frac1{j!}\sum_{i_1+...+i_j=n}\binom{n}{i_1,...,i_j}\beta_{i_1}...\beta_{i_j}$$
where $\beta_n=(-1)^nB_n/n, n\ge1$
So if I rewrite the generating function for the generalized Bernoulli numbers
$$\sum_{n=0}^\infty B_n^{(-m)}\frac{x^n}{n!}=\left(\frac{e^x-1}{x}\right)^m=\exp\left(m \log\left(\frac{e^x-1}{x}\right)\right)=\sum_{j=0}^\infty\frac{m^j}{j!}\left[\log\left(\frac{e^x-1}{x}\right)\right]^j$$
now, since
$$\frac{d}{dx}\log\left(\frac{e^x-1}{x}\right)=\frac{1}{x}\left(\frac{-x}{e^{-x}-1}-1\right)=\sum_{n=1}^\infty{(-1)^nB_n}\frac{x^{n-1}}{n!}$$ where
$$\frac{x}{e^x-1}=\sum_{n=0}^\infty{B_n\frac{x^n}{n!}}$$
we then get that
$$\log\left(\frac{e^x-1}{x}\right)=\sum_{n=1}^\infty{(-1)^n\frac{B_n}{n}}\frac{x^{n}}{n!}=\sum_{n=1}^\infty{\beta_n\frac{x^n}{n!}}$$
and thus
$$\frac{1}{j!}\left[\log\left(\frac{e^x-1}{x}\right)\right]^j=\sum_{n=0}^\infty{\beta_n^{(j)}\frac{x^n}{n!}}$$
Therefore
$$\sum_{n=0}^\infty B_n^{(-m)}\frac{x^n}{n!}=\sum_{j=0}^\infty{m^j}\sum_{n=0}^\infty{\beta_n^{(j)}\frac{x^n}{n!}}=\sum_{n=0}^\infty\sum_{j=0}^\infty{\beta_n^{(j)}{m^j}\frac{x^n}{n!}}$$
So
$$B_n^{(-m)}=\sum_{j=0}^\infty{\beta_n^{(j)}{m^j}}$$
How do I get the finite upper bound $n$? I understand that the lower bound must start at 1 by the condition imposed earlier. I'm sure its silly but I'm stuck.
