This is dealing with the binomial coefficients--namely,$\hspace{4 pt}f(n, k)$ is the number of subsets made of $k$ of the elements of the set $\{1, 2, ..., n\}$. $B_n(x)$ is defined as $$B_{n}(x)=\sum_{k \geq 0} f(n, k)x^k, $$ for $n \in \mathbb{N}=\{0, 1, 2, 3, ...\}.$
On page 14 of the 2nd edition (about 2/3 of the way down the page), it says that by multiplying $$f(n, k)=f(n-1, k)+f(n-1, k-1) \hspace{1 cm} (f(n, 0)=1)$$ throughout by $x^k$ and summing over $k \geq 1$, they obtain $B_n(x)-1=(B_{n-1}(x)-1)+xB_{n-1}(x)$ for $n \geq 1$, with $B_0(x)=1$.
How is $B_0(x)=1$ obtained?
Is it from knowing that the number of size $k$ subsets of a $0$-element set is $0$ when $k>0$, hence making $0 $ the coefficient of $x^k$ in $B_0$(x) when $k>0$ (this was all I could come up with), or is this a consequence of something done in the process of "multiplying throughout by $x^k$ and summing over $k \geq 1$"? Or something else?
Just note that $f(0,0) = 1$, because you have one subset of $\varnothing$ with zero elements: $\varnothing$ itself. And since $\varnothing$ does not have any subset with more than $0$ elements, it follows that $f(0,k) = 0$ for $k \geq 1$. Hence $$B_0(x) = \sum_{k \geq 0}f(0,k)x^k = f(0,0) + f(0,1)x+\cdots = 1 + 0\cdot x + \cdots = 1.$$