mathschallenge.net

Problem

Given that n is a positive integer, prove that (2ⁿ)! is divisible by 2^{2n 1}

Solution

Firstly we write, 2^{2n 1} = 2^{n + n 1} = 2ⁿ 2^{n 1}

Then, (2ⁿ)! = 1 2 3 ... (2ⁿ 1) 2ⁿ

Clearly, (2ⁿ)! is divisible by 2ⁿ, but as 2^{n 1} 2ⁿ, one of the earlier factors of (2ⁿ)! must be 2^{n 1}.

Hence (2ⁿ)! is divisible by 2^{2n 1}.