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:
$1+2+4+...+2048=1+2+4+...+2^{11}=2^{12}-1=4096-1=4095$
$\Rightarrow (1+2)(1+2^2)(1+2^4)...(1+2^{2048})=1+2+2^2+2^3+...+2^{4095}=2^{4096}-1$ $\blacksquare$