Problem
NT-B2-M05-P011 Why \(2^n\) Is Too Large
#11
★★★★☆ Level 4 of 5
Prove that \(2^n\nmid n!\) for every positive integer \(n\).
Bound \(v_2(n!)\) above by a geometric sum.
By Legendre's formula, \(v_2(n!)=\lfloor n/2\rfloor+\lfloor n/4\rfloor+\cdots\). Each term is at most the corresponding \(n/2^i\), and the sum of floors is strictly less than \(n/2+n/4+\cdots=n\). Thus \(v_2(n!)
A proof problem using an estimate, not computation.