I'm studying problem 2.6 (p. 65) in Herbert Wilf's generatingfunctionology (released by the author for free online). This problem actually has a solution already written in the back of the book (p. 203) which I'm trying to understand.
Problem: For $k, n \in \mathbb{N}$, let $f(n,k)$ be defined as follows:
$$ \sum_{n_1 +n_2 + \ldots + n_k = n} n_1 n_2 \cdot \ldots \cdot n_k $$
Find the following:
(i) the ordinary power series generating function of $f$ (i.e., the $opsgf$ of $f$)
(ii) a simple formula for it
Book Solution:
$$ \sum_{n_1 +n_2 + \ldots + n_k = n} n_1 n_2 \cdot \ldots \cdot n_k \\ = [x^n] \left(\sum_r rx^r \right)^k \\ = [x^n] \left \{ \frac{x}{(1-x)^2} \right \}^3 \\ = [x^n] \frac{x^k}{(1-x)^{2k}} \\ $$
Hence $\sum_n f(n,k) x^n = \frac{x^k}{(1-x)^{2k}}$. Explicitly, since
$$ \frac{1}{(1-x)^{2k}} = \sum_{r \ge 0} {r + 2k - 1 \choose r} x^r, $$
we find that $f(n,k) = {n+k-1 \choose n-k}$.
Question: I'm at a loss why the first jump in the equality above holds. In particular, why do we start with
$$ \sum_{n_1 +n_2 + \ldots + n_k = n} n_1 n_2 \cdot \ldots \cdot n_k \\ = [x^n] \left(\sum_r rx^r \right)^k $$
as above? I think I understand this is saying that $f(n,k)$ is the $nth$ term of the power series expansion of $\left(\sum_r rx^r\right)^k$. Why are we raising $\sum_r rx^r$ to the $k$th power? What fundamentally is going on here?
For the first step
Every $n_i$ can have a value $1,2,..$ so the generating function for this $n_i$ is $\sum_r rx^r$. So look at $$ (\sum_r rx^r)^k $$If you look to the coefficient of $x^n$ then you have $n_1+...+n_k=n$ because of the exponents and as coefficients it has $$ \sum_{n_1+...n_k=n}n_1\cdot...\cdot n_k $$ I hope this makes it understandable.