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Is [(2n-1)(2n-2)(2n-22)(2n-23)···(2n-2n-1)] ÷ n! an integer? Why or why not?

My process

Reversed order of numerator and factored out 2's to get:

2(n-1)*(2-1)*2(n-2)*(22-1)....22*(2(n-2)-1)*2*(2(n-1)-1)*(2n-1).

=2(n^2-n)*1*3*7*15*31*63*127*255*511*1023*2047...(2n-1)

Factoring out the 2's from the denominator, we will have a maximum of 2(n-1) so we only need worry about the odd factors in n!:

n! with 2's factored out =1*3*5*3*7*9*5*11 ....

If n=11, then 11 must be a factor of the numerator for the results to be an integer but this is not the case so this does not always result in an integer.

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1023 is evenly divisible by 11.

i suspect its the case that for every prime number p;

2^(p-1)-1 is evenly divisible by p.

can anyone prove this?

Thank you for correcting me. When I first started writing my response, I meant it as only a beginning then thought I saw one that didn't work. I acted too quickly without checking every possibility. I will try to be more careful next time.

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