The book that I'm reading is omitting a lot of steps. I think that i have boiled a step down to this equality. I have tried to prove it by induction, but for some reason I can't make ends meet. Assume that $b_i \neq b_j$ for $i\neq j$ and $b_{i}>0 \ \forall i \in \mathbb{N}$
$\sum_{i=0}^n\frac{\Pi_{j=0,j\neq i}^n(b_j-b_{n+1})}{\Pi_{j=0,j\neq i}^n(b_j-b_{i})}=1$
What is the polynomial having degree $\leq n$ which equals one at $b_0,b_1,\ldots,b_n$? Of course, the polynomial which constantly equals one. For any $j\in\{0,\ldots,n\}$, let $p_j(x)$ be the polynomial which equals $1$ at $x=b_j$ and $0$ at $x=b_k$ with $k\neq j$. We have $$ p_j(x) = \frac{\prod_{k\neq j}(x-b_k)}{\prod_{k\neq j}(b_j-b_k)},\qquad \deg(p_j)=n, $$ and by interpolation $$ \forall x\in\mathbb{R},\qquad \sum_{j=0}^{n}p_j(x) = 1.$$