$\int_0^1\int_0^{\infty}t^{c-1}e^{-(s-z)t}dtz^{a-1}(1-z)^{c-a-1}dz$ = $\int_0^1\frac{1}{(s-z)^c}\int_0^{\infty}t^{c-1}e^{-t}dt z^{a-1}(1-z)^{c-a-1}dz$
I do not understand how we can get from
$\int_0^1\int_0^{\infty}t^{c-1}e^{-(s-z)t}dt$ to $\int_0^1\frac{1}{(s-z)^c}\int_0^{\infty}t^{c-1}e^{-t}dt z^{a-1}(1-z)^{c-a-1}dz$
Specifically, how do we get $\frac{1}{(s-z)^c}$ ?
You make a linear change of variables $\tau = (s-z) t$: $$ \int_0^\infty t^{c-1} e^{-(s-z) t } dt = \int_0^\infty \left(\frac{\tau}{s-z} \right)^{c-1} e^{-\tau} \frac{1}{s-z} d \tau=\frac{1}{(s-z)^{c-1+1}} \int_0^\infty \tau^{c-1} e^{-\tau} d \tau$$