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The order of f(z) is the infimum of all m such that f(z) = O(exp(|z|^m) as z → ∞.

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Berndt showed that the number of zeros of $\zeta^{(k)}(s)$ for $0<t<T$ is $$ N_k(T)=\frac{T}{2\pi}\left(\log\left(\frac{T}{4\pi}\right)-1\right)+O\left(\log T\right). $$ From this I think it follows that $\zeta^{(k)}(s)$ has the same genus and hence the same order as $\zeta(s)$.

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