Prove that $\frac{1}{(1-x)^k}$ is a generating function for $\binom{n-k-1}{k-1}$ On my discrete math lecture there was a fact that:
$\frac{1}{(1-x)^k}$ is a generating function for $a_n=\binom{n-k-1}{k-1}$
I'm interested in combinatorial proof of this fact. Is there any simple proof of such kind available somewhere? (I saw proof using $k$ x $n$ board where we're thinking of $n-k-1$ roads availible to point $(k,n)$ where we're choosisng $k-1$ fields to go right but I didn't understand it - is there a simpler proof somewhere?)

Suppose we have a sequence of values of length $k$ where the values are any non-negative integer. This is has the combinatorial specification
$$\mathfrak{S}_{=k}\left(\sum_{q\ge 0} \mathcal{Z}^q\right).$$
This gives the generating function
$$\frac{1}{(1-z)^k}.$$
On the other hand all such sequences can be obtained by stars-and-bars combining $n$ elements with $k-1$ dividers, giving
$$\sum_{n\ge 0} {n+k-1\choose k-1} z^n,$$
which was to be shown.
