Com10atorics wrote: Find the value of $(1+2)(1+2^2)(1+2^4)(1+2^8)...(1+2^{2048})$. $2^{4096}-1$

Multiply this expression by $(2-1)$ and you obtain by difference of squares $2^{4096}-1$ This should be in MSM edit: sniped by 2 secs

Notice that: $(1+2)(1+2^2)=1+2+2^2+2^3$ $(1+2)(1+2^2)(1+2^4)=1+2+2^2+2^3+2^4+2^5+2^6+2^7$ And so on $(1+2)(1+2^2)(1+2^4)...(1+2^k)=1+2+2^2+2^3+...+2^{1+2+4+..+k}$, where k is a power of 2 In this case $k=2048$, so: $1+2+4+...+2048=1+2+4+...+2^{11}=2^{12}-1=4096-1=4095$ $\Rightarrow (1+2)(1+2^2)(1+2^4)...(1+2^{2048})=1+2+2^2+2^3+...+2^{4095}=2^{4096}-1$ $\blacksquare$