Given the following function:
$f(1) = 2$
$f(x+1) = 2^{f(x)}$
Show that $f(i) > f(i-1)^{i-1}$. Starting with some $i > i_0$.
Intuitively I can easily see why this is so. Basically $f(x) = 2^{2^{2^{2^{2\dots}}}}$ where there is an $x$ number of $2s$, while $f(x-1)^{x-1} = (2^{2^{2^{2\dots}}})^{x-1}$ where there are $x-1$ number of $2s$. So in the first case the exponent grows, while in the second the base grows.
I have tried to show this using induction.
So far I have: IH: $f(i + 1) > f(i)^i$.
$f(i+2) = 2^{f(i + 1)} = 2^{2^{f(i)}}$
$f(i + 1)^{i + 1} = (2^{f(i)})^{i + 1} $
I really cannot see how to apply the IH to show my inequality. I have been stuck on this for a really really long time, I would very much appreciate any help.