Transfinite Knuth-arrow hierarchy vs. fast-growing hierarchy Suppose Knuth arrow notation (and hence the hyperoperation sequence) is extended to transfinite ordinal indices as follows:
Let μ be a large countable ordinal such that a fundamental sequence is assigned to every limit ordinal less than μ, and take these to be the same as in the definition of the fast-growing hierarchy of functions ($f_\alpha$). Now define
$$
  m\uparrow^\alpha n \ \ = \ \
  \begin{cases}
    m n & \text{if }\alpha=0, \\
    (m\uparrow^{\alpha-1})^n 1 & \text{if }\alpha \text{ is a successor ordinal}, \\
    m\uparrow^{\alpha[n]}n & \text{if }\alpha \text{ is a limit ordinal},
  \end{cases}
$$
for all integers $m, n \ge 0$ and ordinal $\alpha < \mu$, where  $(\alpha[n])_{n=0,1,2,\dots}$ is the fundamental sequence for $\alpha$.
It can be shown that for all $\alpha < \omega$, there is an $n_0$ (depending on $\alpha$) such that for all $n > n_0$, 
$$f_\alpha(n) \ \ \lt \ \ 2\uparrow^\alpha n \ \ \le \ \ f_{\alpha+1}(n).\tag{*}$$
Question: Does (*) hold for all $\alpha < \mu$? Or, is there a "cross-over point", i.e., a least ordinal $\beta (< \mu)$ such that for all $\alpha \ge \beta, \ \ \ f_\alpha(n) > 2\uparrow^\alpha n \ $ for all sufficiently large $n$? 
EDIT: Small cases $n = 0, 1, 2, 3, ...$ suggest that $\omega$ is actually the cross-over point. Proof?
Motivation: If (*) holds for all $\alpha < \mu$, it would seem that with minor adjustments virtually all of the points of interest concerning the fast-growing hierarchy ($f_\alpha)$ apply as well to the hierarchy of functions $(2\uparrow^\alpha)$.
NB: Arrow notation has been used for convenience. The discussion could be rephrased in terms of extending the hyperoperation sequence $H_\alpha(m,n)$, with a shift in the indexing to accomodate successor and addition.  

Certainly this eventual inequality holds for all $\alpha<\omega$. Then note that
$$f_\omega(n)=f_{\omega[n]}(n)\ll2\uparrow^{\omega[n]}n=2\uparrow^\omega n$$
So the lower bound holds for $\omega$. Likewise does the upper bound somewhat trivially. However, it fails for $\omega+1$. Let $\omega[n]=n$. Note that
$$f_\omega(n+1)=f_{n+1}(n+1)\gg2\uparrow^n(n+1)\ge(2\uparrow^nn)+1=(2\uparrow^\omega n)+1$$
Hence by induction,
$$f_\omega^k(n+1)\gg((2\uparrow^\omega)^kn)+1>(2\uparrow^\omega)^k1$$

$$f_{\omega+1}(n)=f_\omega^n(n)\gg(2\uparrow^\omega)^n1=2\uparrow^{\omega+1}n$$

