By plotting the function and its derivatives, one can easily be convinced that the function $$f(x):=\log\binom{x}{p x}=\log\Gamma(x+1)-\log\Gamma(px+1)-\log\Gamma((1-p)x+1),$$ defined for $x>0$ and $p \in (0,1)$, is completely monotone (i.e., for all $x$, $f(x)>0$, $f'(x)<0$, $f''(x)>0$, etc). How can this statement be proved, or is there a known proof?

By Bernstein's theorem on monotone functions, the statement is equivalent to $f$ being the Laplace transform of a non-negative function. I tried to use the Taylor expansion of log-gamma $$\log\Gamma(x+1) = -\gamma x+\sum_{j=2}^\infty \frac{\zeta(j)}{j} (-x)^j,$$ and divide the $j$th term by $j!$ (which is basically what the inverse Laplace transform does). But the result doesn't seem to be a known function (without the $\zeta(j)$ constants, though, the resulting series would essentially become the expansion of the exponential integral function.

Edit: The fact that $f'(x)<0$ is equivalent to concavity of the digamma function, which can be shown by considering its integral form (see the first answer below). However, the higher derivatives seem less obvious to work with.