I have the following problem. I need to show that $b(i)>b(i-1)^{i-1}$ for $i>k$ for some $k$. $b$ is the following function: $b(1)=2$ and $b(n)=2^{b(n-1)}$
I tried to do this by induction, but it does not work out: Take base case $k=2$. Then it works.
Induction hypothesis: for $n$ we have that $b(n)>b(n-1)^{n-1}$. Then for $b(n+1): b(n+1)=2^{b(n)}$ From our IH we have that $2^{b(n)} > 2^{b(n-1)^{n-1}}$ We can rewrite the right term to $b(n)^{n-1}$. Now I end up with: $b(n+1)>b(n)^{n-1}$, whereas I want $b(n+1)>b(n)^n$.
Does anybody have any tips of how to prove this?
Observe that $b(n) \geqslant n+1$ hence $$b(n+1) =2^{b(n)} >2^{b(n-1)^{n-1}} =(2^{b(n-1)})^{b(n-1)^{n-2}} =b(n)^{b(n-1)^{n-2}}\geqslant b(n)^{b(n-1)} \geqslant b(n)^n$$ for $n\geqslant 3.$