Related with The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts
I know that ${(x^n-1)\over (x-1)}=1+x+x^2+x^3+x^4+...$ as a Geometric Progress, but I do not see how $${1\over (1-x)(1-x^3)(1-x^5)\cdot\dots}=(1+x+x^{1+1}+\dots)(1+x^3+x^{3+3}+\dots)(1+x^5+\dots).$$ can anyone please explain?