Not really an answer to your question, but the expression you've post is the way of generating the partitions of integers.
Write out the first few terms, [tex](1+x)(1+x^{2})(1+x^{3})(1+x^{4})(1+x^{5})(1+x^{6})(1+x^{7})[/tex]
You can see that the coefficient of the constant term in the expansion will be just 1.
coefficient of the x term will be 1
coefficient of the x^2 term will be 1
coefficient of the x^3 term will be 2, because x^3 can be made from [itex]x.x^{2}[/itex] or just x^3 (ie 3 = 3+0 or 2+1)
x^4 will have coefficient 2 because x^4 comes from x^4 and x^3 * x (ie 4 = 4+0 or 3+1)
x^5 is 3, because 5 = 5+0 or 5=4+1 or 5=3+2
x^6 is 4 because 6=6+0 or 6=5+1 or 6=4+2 or 6 = 3+2+1)
See how it works? The coefficient of x^n will be the number of ways you can express n as a sum of unique integers. The closed form for this I don't think exists.