Prove these two functions have the same coefficients We have $\displaystyle p(x) = \frac{1}{1-x}\cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^5} \cdot \ ...$ and $\displaystyle q(x) = (1+x)\cdot(1+x^2) \cdot (1+x^3)\cdot \ ...$. 
Let's say that these two functions have coefficients $a_n$ and $b_n$ respectively.
Now I have to prove that $a_n = b_n$ for all $n$.
I know that $\displaystyle \frac{1}{1-x} = \sum_{n=0} ^\infty x^n$ and  $\displaystyle \frac{1}{1-x^3} = \sum_{n=0} ^\infty x^{3n}$ etc.
But how about the terms of $q(x)$? And how can I get one power series for $p(x)$ and $q(x)$? 
 A: The coefficient of $x^n$ in the product
$$ (1+x)(1+x^2)(1+x^3)\cdot\ldots $$
gives the number of ways of writing $n$ as a sum of distinct natural numbers, while the coefficient of $x^n$ in the product:
$$ \frac{1}{1-x}\cdot\frac{1}{1-x^3}\cdot\ldots = (1+x+x^2+\ldots)(1+x^3+x^6+\ldots)\cdot\ldots $$
gives the number of ways to write $n$ as a sum of different natural numbers, where every number occurs an odd number of times. Now have a look at the paragraph "Odd parts and Distinct parts" in the following Wikipedia page. We have:
$$\prod_{n\geq 1}(1+x^n) = \prod_{n\geq 1}\frac{1-x^{2n}}{1-x^n}=\frac{\prod_{n\geq 1}(1-x^{2n})}{\prod_{n\geq 1}(1-x^{2n})\prod_{n\geq 1}(1-x^{2n-1})}=\prod_{n\geq 1}\frac{1}{1-x^{2n-1}}.$$
This is just a special case of Glaisher's theorem.
A: Hint:
Developing the two expressions with just enough terms to get the exact $x^7$ coefficient,
$$\frac1{1-x}\frac1{1-x^3}\frac1{1-x^5}\frac1{1-x^7}\\
=(1+x+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^3+x^6)(1+x^5)(1+x^7)\\
=\color{green}1+\color{green}1x+\color{green}1x^2+\color{green}2x^3+\color{green}2x^4+\color{green}3x^5+\color{green}4x^6+\color{green}5x^7\cdots$$
and
$$(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)(1+x^7)\\
=\color{green}1+\color{green}1x+\color{green}1x^2+\color{green}2x^3+\color{green}2x^4+\color{green}3x^5+\color{green}4x^6+\color{green}5x^7\cdots$$
This cannot be a coincidence.
