You only need to worry about when n(2n-1) is not of the form 2ⁿ5^m.

2n-1 is always odd, so that must always be a power of 5.

2n-1=5^k ==> n=(5^k-1)/2

So then you need (5^k-1)/2 to be factored as powers of 2 and 5. 5 will never be a factor, so we need (5^k-1)/2=2^m ==> 5^k-1=2^m+1

This is true for k=1, which is n=2.

All other cases are when (4N+1) happens to cancel any non-2 and non-5 factors from the denominator, which happens at N=8. (2n-1 = 15, 4n+1=33, so the 3 cancels leaving 2^3*5 in the denominator).

Oops, yeah.

Yep, that is correct. 5^k+1 mod 100 = 26, so  (5^k+1)/2 mod 10 = 3. Then (5^k+1)/2=5^m is never true, so the only way to take care of that is if (5^k+1)/2=(4n+1)2^r, but since it is odd, we know it must be that r=0, so (5^k+1)/2=(4n+1)

Also, sorry if I am wrong about anything, I haven't slept much.