It is conceivable that ways do exist to express $\log (\lfloor x\rfloor !)$ in an expression involving the zeros, but any apparent relationship will be superfluous because the "explicit formula" for this function comes from $\zeta'(s)$, not something involving $1/\zeta(s)$.

When $x$ is not an integer, you have

$$\log (\lfloor x\rfloor !)=\sum_{n\leq x}\log n=-\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\frac{\zeta'(s)x^sds}{s}.$$

The double pole at $s=1$ gives you a residue of $x\log x-x$ and the integrand has no other poles in $\sigma>0$. Thus, you have

$$\log (\lfloor x\rfloor !)=x\log x-x-\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{\zeta'(s)x^sds}{s}$$

for any $0<\sigma<1$. (You can't cross the imaginary axis because the integral doesn't converge when $\sigma<0$. One way to see this is to consider Stirling's approximation $$\log ((x-1)!)=(x-1/2)\log x-x +(1/2)\log 2\pi +o(1)$$

because, if it were valid to cross the imaginary axis, picking up an $O(1)$ contribution from the residue of the pole at $s=0$, we would have a contradiction).

To evaluate the $O(\log x)$ terms coming from the integral, you have to do a little more work. I think Binet gave an alternative expression as a Laplace transform, and there are both convergent (Macdonald) and asymptotic (Stirling) series expansions. Try wikipedia as a starting point.

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