This is a bit silly but I am unable to arrive at the conclusion to example 2.8. This is the number of ways that a number can be written as an ordered sum. Wilf shows that
$$f(n,k) = \frac{1}{(1-x)^{k}}$$
where $f(n,k)$ is the ordered sums of $n$ containing $k$ terms. Wilf concludes:
By (1.31), $f(n,k) = {n+k-1 \choose n}$
Looking back to 1.31, these are the two main binomial results:
$$ \sum_{k} {n \choose k}x^{k} = (1+x)^n $$
and
$$ \sum_{n} {n \choose k}y^{n} = \frac{y^{k}}{(1-y)^{k+1}} $$
It is not clear, but if I had to guess it is the second result that is relevant. Using this, I get (with some different indices):
$$\frac{x^{p}}{(1-x)^{p+1}} = \sum_{q} {q \choose p}x^{q}$$ $$\frac{1}{(1-x)^{p+1}} = \sum_{q} {q \choose p}x^{q-p} $$
Continuing, I then substitute $k = p+1$:
$$\frac{1}{(1-x)^{k}} = \sum_{q} {q \choose k-1}x^{q-k+1} $$
I can then try to get it to look "like" the result with $n = q - k + 1$:
$$ \frac{1}{(1-x)^{k}} = \sum_{n} {n+k-1 \choose k-1}x^{n} $$
but this is still wrong.
Any guidance would be appreciated.
Note that $\binom{p}{q}=\binom{p}{p-q}$. So we have \begin{align*} \frac{1}{(1-x)^k}=\sum_{n} {n+k-1 \choose k-1}x^{n}=\sum_{n} \color{blue}{{n+k-1 \choose n}}x^{n} \end{align*} according to the claim.