Don't understand one step for AVL tree height log n proof I came across a proof that the an AVL tree has O(log n) height and there's one step which I do not understand.

Let $N_h$ represent minimum number of nodes that can form an AVL tree of height h. Since we're looking for minimum number of nodes, let its children's minimum number of nodes be $N_{h-1}$ and $N_{h-2}$.

Proof:

$$N_h = N_{h-1} + N_{h-2} + 1 \tag{1}$$
$$N_{h-1} = N_{h-2} + N_{h-3} + 1 \tag{2}$$
$$ N_h = (N_{h-2} + N_{h-3} + 1 + ) + N_{h-2} + 1 \tag{3}$$
$$ N_h &gt; 2N_{h-2} \tag{4}$$
$$N_h &gt; 2^{h/2} \tag{5} $$

I do not understand how we went from (4) to (5). If anyone could explain, that'd be great. Thanks!
You can continue as same as line 4 the process like that:

$$ N_h &gt; 2N_{h-2}&gt; 2(2 N_{h-4})&gt;2(2(2 N_{h-6}))&gt;\cdots$$
As you can see, the indexs are decreasing by substracting $2$ in each step when you use the inequality. So, the process stops when the index $h$ takes $0$, but from the indexs behavior the half of $h$ (floor) will be the quantity of times that we substract $2$ from $h$.

$$ N_h &gt; 2N_{h-2}&gt;\cdots&gt;2(2(2(2(2h^{h-10}))))&gt; 2^{\frac{h}{2}}$$