A problem of combinatorics (reduced from a problem of computing probability generating function) This is a problem involving computation of probability generating function (PGF).

I have reduced the problem into a problem of combinatorics, as given below.
Let $f(n,k)$ be defined as
$$f(n+1,k):=f(n,k)+f(n,k-1)+\cdots+f(n,k-n)$$
where $f(n,k)=0$ for $k<0$, $f(1,0)=1$ and $f(1,k)=0$ otherwise. To show that
$$\sum_{k=0}^{\infty}s^k f(n,k)=\prod_{k=2}^n \frac{1-s^k}{1-s},\,\, |s|<1$$
Source : Rohatgi, Saleh. p.92. Problem 11. Thanks in advance.
 A: You can solve this using induction; the base case $n = 1$ follows trivially by definition of $f(1,k)$ as given in the question. 
We assume that the statement is true for a given value of $n$
$$\sum_{k=0}^\infty s^k f(n,k) = \sum_{k=-\infty}^\infty s^k f(n,k) = \prod_{m = 1}^n \frac{1 - s^m}{1-s}.$$ 
Note that the formula you give for $f(n+1,k)$ can be written as
$$f(n+1,k) = \sum_{j=0}^{n} f(n,k-j).$$
Now using the above
\begin{align*}
\sum_{k=-\infty}^\infty s^k f(n+1,k) & = \sum_{k=-\infty}^\infty s^k 
 \sum_{j = 0}^{n} f(n,k - j ) \\
& = \sum_{j=0}^{n} \sum_{k=-\infty}^\infty s^k f(n,k-j) \\
& = \sum_{j=0}^{n} s^{j} \sum_{k=\infty}^\infty s^{k -j} f(n,k-j) \\
& = \sum_{j=0}^{n} s^j \left( \prod_{m=1}^n \frac{1-s^m}{1-s} \right)\\
& = \left( \prod_{m=1}^n \frac{1-s^m}{1-s} \right) \left( \sum_{j=0}^{n} s^j \right) \\
& = \prod_{m=1}^{n+1} \frac{1-s^m}{1-s}.
\end{align*}
A few notes to clarify some of the steps:
Although unncessary, I have used the convention of summing from $k = -\infty$ to $\infty$, rather than $k= 0$ as in the question. This is really just to make the point that we don't have to worry that shifting the series might lead to un-defined series entries; we know $f(n,k) = 0$ for $k \leq 0$ so this is ultimately not important.
In the second line we change the order of summation; this is possible because the terms in the series are all positive (you may feel you want to prove this to yourself).
Going between the fourth and fifth line, I am using the fact that the product is independent of $j$ to bring the product terms outside the summation.
Finally to get from the fifth to sixth lines I use the identity 
$$\sum_{j=0}^n s^j = \frac{1 - s^{n+1}}{1-s}, \qquad |s| < 1.$$
A: The problem can be rewritten as
$$
\left\{ \matrix{
  f(n,k) = \sum\limits_{0\, \le \,j\, \le \,n - 1} {f(n - 1,k - j)}  \hfill \cr 
  f(n,k) = 0\quad \left| {\,n,k < 0} \right. \hfill \cr 
  f(0,k) = f(1,k) = \left[ {0 = k} \right] \hfill \cr}  \right.
$$
and more compactly as
$$
f(n,k) = \sum\limits_{0\, \le \,j\, \le \,n - 1} {f(n - 1,k - j)}  + \left[ {0 = n} \right]\left[ {0 = k} \right]
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$
Then let's take the generating function on the index $k$
$$
\eqalign{
  & F(n,x) = \sum\limits_{0\, \le \,k} {f(n,k)\,x^{\,k} }  = \sum\limits_{0\, \le \,j\, \le \,n - 1} {\sum\limits_{0\, \le \,k} {f(n - 1,k - j)\,x^{\,k} } }  + \left[ {0 = n} \right] =   \cr 
  &  = \sum\limits_{0\, \le \,j\, \le \,n - 1} {x^{\,j} \sum\limits_{0\, \le \,k} {f(n - 1,k - j)\,x^{\,k - j} } }  + \left[ {0 = n} \right] =   \cr 
  &  = {{1 - x^{\,n} } \over {1 - x}}F(n - 1,x) + \left[ {0 = n} \right] \cr} 
$$
Thus
$$
\eqalign{
  & F(0,x) = 1  \cr 
  & F(1,x) = {{1 - x^{\,1} } \over {1 - x}} = 1  \cr 
  & F(2,x) = {{1 - x^{\,2} } \over {1 - x}} = {{1 - x^{\,2} } \over {1 - x}}{{1 - x^{\,1} } \over {1 - x}}  \cr 
  & \quad \quad  \vdots   \cr 
  & F(n,x) = \sum\limits_{0\, \le \,k} {f(n,k)\,x^{\,k} }  = \prod\limits_{1\, \le \,\,k\, \le \,n} {{{1 - x^{\,k} } \over {1 - x}}}  \cr} 
$$
$f(n,k)$ is OEIS sequence A008302.
