We can use the property $\Gamma{(a + 1)} = a \Gamma{(a + 1)}$ to simply the expression below to $\frac{a}{a + b}$ if $k = 1$.
$$ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{\Gamma(a+1)}{\Gamma{(a + 1 + k)}} $$
How does it simplify? Here's what I've tried.
$$ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{\Gamma(a+1)}{\Gamma{(a + 1 + k)}} \\ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{a\Gamma(a)}{\Gamma{(a + 2)}} \\ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{a\Gamma(a)}{a^2\Gamma{(a)}} \\ \frac{a^b\Gamma{(a)}}{\Gamma{(a)}} \frac{a\Gamma(a)}{a^2\Gamma{(a)}} \\ \frac{a^{b+1}}{a^2} \\ a^{b - 1} $$
That doesn't look like $\frac{a}{a + b}$. Where did I err?
My guess is that the formula simplifies to $\frac{a}{a+b}$ when $k=b$ (and not when $k=1$). To see this, use the identities $\Gamma(a+1+b)=(a+b)\Gamma(a+b)$ and $\Gamma(a+1)=a\Gamma(a)$.