Evaluate $ 1 + \sum_{i=0}^{n-2} [ \prod_{k=0}^{i} (n-k)/(r-k) ] $ Is there a way in which the expression
$$ 1 + \sum_{i=0}^{n-2}  \prod_{k=0}^{i} \dfrac{n-k}{r-k}  $$
may be simplified?
It would simplify some equations I am working with.

First, using recurrence equation for Euler's $\Gamma$-function:
$$
   \prod_{k=0}^m \frac{n-k}{r-k} = \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma(r+1)}
$$
Then, changing summation variable as $m \to n-2-m$:
$$
   1 + \sum_{m=0}^{n-2} \frac{\Gamma(n+1)}{\Gamma(n-m)} \cdot \frac{\Gamma(r-m)}{\Gamma(r+1)} = 1 + \sum_{m=0}^{n-2} \frac{\Gamma(n+1)}{\Gamma(n-(n-2-m))} \cdot \frac{\Gamma(r-(n-m-2))}{\Gamma(r+1)} = \\
1 + \frac{\Gamma(n+1)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)} 
 = 1 + \frac{\Gamma(n+1)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)} 
$$
Now the last sum is just a hypergeometric series:
$$
   \frac{\Gamma(n+1)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2)}  = 
   \frac{\Gamma(n+1) \Gamma(1+r-n)}{\Gamma(r+1)}  \sum_{m=0}^{n-2} \frac{\Gamma(m+r+2-n)}{\Gamma(m+2) \Gamma(1+r-n)}  = \\
    \frac{1}{\binom{r}{n}} \sum_{m=0}^{n-2} \binom{m+r-n+1}{m+1} = \frac{n}{r-n+1} - \frac{1}{\binom{r}{n}}
$$
The last equality follows because $s_{m+1} - s_m = \binom{m+r-n+1}{m+1}$, where
$$
    s_m = \frac{m+1}{r-n+1}\binom{m+r-n+1}{m+1}
$$
And therefore the original sum telescopes 
$$ \sum_{m=0}^{n-2} \binom{m+r-n+1}{m+1} = \sum_{m=0}^{n-2} (s_{m+1}-s_m)  = s_{n-1} - s_0
$$
It now remains to compute $s_{m+1}-s_m$:
$$
  s_{m+1} = \frac{m+2}{r-n+1} \binom{m+r-n+2}{m+2} = \frac{m+2}{r-n+1} \cdot \frac{m+r-n+2}{m+2} \binom{m+r-n+1}{m+1} = \\ s_m + \binom{m+r-n+1}{m+1}
$$
