When a body falls from a height h, it strikes the ground with a velocity $u = \sqrt{2gh}$. Let it rebounds with a velocity v and rise to a height $h_1$
Therefore,$v=\sqrt{2gh_1}$
$e=\frac{v}{u}=\sqrt{\left(\frac{h_1}{h}\right)}$
If the successive heights to which the body rebounds again and again and $h_2, h_3$,..., then
$e=\sqrt{\frac{h_1}{h}}=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{h_1}{h_2}}=...$
Clearly, $h_1, = e^2h, h_2 = e^4h$ ... and so on.
Similarly, after nth rebound
$h=e^{2n}h$