Given a list of length n, how many times can it be bisected before all the pieces are of length one?

After p cuts the length of each piece is n/(2p). We are looking for p such that ( n/(2p) ) 1 and ( n/(2p-1) ) > 1, i.e.

n 2p and n > 2p-1 :2p n>2p-1

p log2n > p-1

The function giving the number of steps in terms of the length n is(log2n)

How to prove this?