Define the Bernoulli numbers $B_n$, by the power series $$ \dfrac{z}{e^z-1}=\sum_{n=0}^\infty \frac{B_n}{n!}z^n. $$ Now prove the following recursion formula $$ \frac{B_0}{n!0!}+\frac{B_1}{(n-1)!1!}+...+\frac{B_{n-1}}{1!(n-1)!} =\begin{cases}1 & \text{if }n=1 \\ 0 & \text{if } n>1\end{cases} \tag{1}$$
I'm having trouble proving this formula. First I noticed that (1) can be rewritten as $$ \frac{1}{n!}\sum_{k=0}^{n-1}{n \choose k} B_k =\begin{cases}1 & \text{if }n=1 \\ 0 & \text{if } n>1\end{cases} \tag{2}$$ I attempted to prove this through induction. The $n=1$ and $n=2$ bases cases can be verified by directly computing the series expansion of $\dfrac{z}{e^z-1}$. So inductively I assume that (2) holds for $n=j$ which gives us $$\frac{1}{j!}\sum_{k=0}^{j-1}{j \choose k} B_k =0 \tag{3} $$ and this implies that $$ \sum_{k=0}^{j-1}{j \choose k} B_k =0 \tag{4} $$ as well as $$ \sum_{k=0}^{j}{j \choose k} B_k =B_j \tag{4} $$ Now I want to show that (2) holds for $n=j+1$, or in other words $$\sum_{k=0}^{j}{j+1 \choose k} B_k =0 \tag{5} $$ I tried applying the formula $ {{n+1} \choose {k}} = {n \choose {k-1}} + {n \choose k}$ to (5) which then gives us $$\begin{align} \sum_{k=0}^{j}{j+1 \choose k} B_k &=\sum_{k=0}^{j}\left({j \choose {k-1}} + {j \choose k}\right)B_k \\&= B_j + \sum_{k=0}^{j}{j \choose {k-1}} B_k \tag{6} \end{align} $$ where in (6) we used (4). But I haven't been able to show that the sum in (6) is equal to $-B_j$.