User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-28606698-20170922185938

1) I mean

$$f_{\omega+1}(10)=f_{f_\omega^9(10)}(f_\omega^9(10))\approx f_{10\rightarrow10\rightarrow9\rightarrow2}(10)$$ and hence for calculation of $$f_{\omega+1}(10)$$ we should come through $$\approx10\rightarrow10\rightarrow9\rightarrow2$$ successor ordinals from $$f_0(f_\omega^9(10))$$ and up to $$f_{f_\omega^9(10)}(f_\omega^9(10))$$ since $$f_\omega^9(10)\approx10\rightarrow10\rightarrow9\rightarrow2$$.

2) You defined

$$h_0(n)=10\uparrow^n10 \approx f_\omega(n)$$

$$h_{k+1}(n)=h_k^n(n)$$

Then $$h_{90}(10)\approx f_{\omega+90}(n)$$ and that is much smaller than $$f_{\omega2+1}(10)\approx f_{\omega+(10\rightarrow10\rightarrow10\rightarrow10\rightarrow2)}(10)$$