Who is going to expand?

We must change the equation a little bit without changing its sense.

$(x-1)(2x - 1)(4x-1)....(1- {2}^{101}x)$

Now, this is a 102 -degree polynomial with roots as - $\frac{1}{{2}^{n}}$ for $n \in [0,101]$

Now, we need to find the co-efficient of ${x}^{101}$ .

By vieta,

Sum of roots = $\frac{-b}{a}$

Hence, by using geometric sum formula,

$\frac{-(Co-efficient of {x}^{101})}{(Co-efficient of {x}^{102})} = \frac{{2}^{102} - 1}{{2}^{101}}$

Now, Co-efficient of ${x}^{102}$ will just be the product of all powers of 2 or ${2}^{0 + 1 + 2 + 3 + 4 + .... 101}$ or just ${2}^{\frac{102*101}{2}}$

After some bashing,

$-(Co-efficient of {x}^{101}) = {2}^{5050}({2}^{102} - 1)$

which is just -

$Co-efficient of {x}^{101} = {2}^{5050} - {2}^{5152}$

Even better, notice that there are 102 choose 101 = 102 ways to get a product of factors such that you will get a x^101 terms.So all you have to do is find the sum of 2^1+2^2 + .... + 2^102 + 1 2^2 + .... + 2^102 +... + 1 2^1*2^2 + .... + 2^101 wherein each term of this sum you omit one of the 2^k terms where k is between 0 and 102 inclusive.