Let $n \to \infty$ and $r \to \infty$, so that the average number of particles per cell, $r/n$, tends to $\lambda$. Then,
$$ {n + r - k -2 \choose r-k} / {n+r-1 \choose r} \to \frac{\lambda^k}{(1+\lambda)^{k+1}}.$$
$ {n + r - k -2 \choose r-k} / {n+r-1 \choose r} = \frac{(n+r-k-2)!r!(n-1)!}{(r-k)!(n-2)! (n+r-1)!} = \frac{r(r-1) ... (r-k+1) (n-1)}{(n+r-1)(n+r-2) ... (n+r-k-1)}$.
Numerator and denominator have both $k+1$ terms. I guess that I need to manipulate the right hand side in some way to use "$r/n \to \lambda$". But I can't figure it out. Can you give some hint?
Hint: Notice that in your last expression there are k+1 terms in the numerator and $k+1$ terms in the denominator$$\frac{r(r-1)\cdots (r-k+1) (n-1)}{(n+r-1)(n+r-2) \cdots (n+r-k-1)}=\color{red}{\frac{n^{k+1}}{n^{k+1}}}\frac{r(r-1)\cdots (r-k+1) (n-1)}{(n+r-1)(n+r-2) \cdots (n+r-k-1)}=\frac{\frac{r}{n}(\frac{r-1}{n})\cdots (\frac{r-k+1}{n}) (\frac{n-1}{n})}{(\frac{n+r-1}{n})(\frac{n+r-2}{n}) \cdots (\frac{n+r-k-1}{n})}.$$ Now take the limit.