A bit of plotting suggests that $\zeta^{(k)}(s) < 0$ for all $s\in (0,1)$ and all integers $k\geq 0$. (Or, what is the same: $\zeta^{(k)}(s)$ has no zeroes on $(0,1)$.) Is there a brief, clean proof of this apparent fact (and/or a reference for it)?



$\zeta^{(k)}(s) < 0$ for $s\in (0,1)$
The coefficients computed in the comments appear to imply that the Taylor expansion at $s=0$ of $\zeta(s)+\frac1{1-s}-\frac12$ has very small coefficients, which would imply the result.


Following section 2.1 in Titchmarsh Theory of the Riemann zeta function, 
by integration/summation by parts (or one step of Euler-Maclaurin),
$$\zeta(s)=\frac1{s-1}+\frac12+s\int\_1^{\infty}\frac{1/2-\{x\}}{x^{s+1}}dx$$
absolutely convergent for $\Re(s)>0$ where $\{x\}$ denotes the fractional part of $x$.


Integrating by parts again (or two steps of Euler-Maclaurin),
$$\zeta(s)=\frac1{s-1}+\frac12+\frac{s}{12}-(s+1)\int\_1^{\infty}(\{x\}^2-\{x\}+\frac16)x^{-s-2}dx$$


Consider the integrand $(\{x\}^2-\{x\}+\frac16)x^{-s-2}dx$ as a function of $s$. Its Taylor coefficients are alternating in sign and dominated by the coefficients of $\frac16x^{-s-2}$, so the Taylor coefficients of the absolutely convergent integral are dominated by those of
$$\int\_1^{\infty}\frac16x^{-s-2}dx = \frac1{6(s+1)}$$
and the result follows.


