how to Prove that 2n!/2^n is an integer ?

Question : Prove that 2n!/2^n is an integer Answer : To Prove that [\frac{2n!}{2^n}] is an integer 2n!2n Using the definition of the factorial, [=>\frac{2n!}{2^n}=\frac{((2n)(2n-1)(2n-2)(2n-4)………3.2.1}{2^n}] Writing it as [=>\frac{2n!}{2^n}=\frac{2.4.6……(2n-4)(2n-2)1.3.5…(2n-3)(2n-1)}{2^n}] [\frac{2n!}{2^n}=\frac{2^n(1.2.3…….n)(1.3.5…(2n-3)(2n-1))}{2^n}] [=>\frac{2n!}{2^n}=(1.2.3…….n)(1.3.5…(2n-3)(2n-1))] [=>\frac{2n!}{2^n}=n!(1.3.5…(2n-3)(2n-1))] Since factorial of a number is an integer…

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