If 1, ω₁ , ω₂ , ......, ω₉₉ be the 100th roots of unity, then (2 – ω) (2 – ω₂) (2 – ω₃) .... (2 – ω₉₉) is

Consider 2^ndroot of unity
x²-1=0
x=+1,-1

Now, forming an equation using these roots
(x-1)(x+1)
=x²-1

Now, Consider 4^throot of unity
x⁴-1=0
(x²+1)(x²-1)=0
x= +i,-i,+1,-1

Now, if an equation is formed using these roots altogether,

(x+1)(x-1)(x+i)(x-i)=(x²+1)(x²-1)= x⁴-1

which is the root equation from which these roots were derived.
Hence,

for the 100^throot, conversely
(x–ω)(x–ω₂)(x–ω₃)....(x–ω₉₉)= x¹⁰⁰-1 ...(asω_{1,2,3,........99}are again roots of unity)
Substituting x=2, we get,
2¹⁰⁰-1