Showing that $B_0=1$ and that $\sum^n_{k=0}{{n+1}\choose k}B_k =0$

Question

Let $h(x)=x$ and $f(x)=e^x-1$ for $x\in \mathbb{R}$. It's known that $\frac{h(x)}{f(x)}$ has a Taylor series in $0$, which is written as follow:

$$\sum^\infty_{n=0}\frac{B_n}{n!}x^n$$

So, how do I show that $B_0=1$ and that $$\sum^n_{k=0}{{n+1}\choose k}B_k =0$$

Answer 1

$$1=\frac{x}{e^x-1}\frac{e^x-1}{x}=(\sum_{n=0}^\infty\frac{B_n}{n!}x^n)(\sum_{n=0}^\infty\frac{x^n}{(n+1)!})=\sum_{n=0}^\infty(\sum_{k=0}^n\frac{1}{(n+1-k)!}\frac{B_k}{k!} )x^n$$ $$1=\sum_{n=0}^\infty(\sum_{k=0}^n{{n+1}\choose k}{B_k})\frac{x^n}{n!} $$ $$\delta_{n,0}=\sum_{k=0}^n{{n+1}\choose k}{B_k}$$