Question regarding: $\prod\limits_{n=1}^{\infty}(1-x^n) = \prod\limits_{n=1}^{\infty} (1-x^{3n})(1-x^{3n-1})(1-x^{3n-2})$ I am studying analytic number theory from Tom M. Apostol and I cannot think about this argument 
related to infinite products, which is to be used in a proof. 

I know that every positive integer can be represented as one among $3n-1$, $3n-2$, $3n$, so can it be deduced that $$\prod_{n=1}^{\infty}  \left(1 - x^n\right) = \prod_{n=1}^{\infty} \left(1-x^{3n}\right)\left(1-x^{3n -1}\right) \left(1-x^{3n-2}\right) .$$

If so, can someone please give a rigorous argument?
 A: If a sequence converges to a value, so does any subsequence. The LHS is by definition equal to $$\lim_{n\to\infty}\prod_{k=1}^n\left(1-x^k\right).$$ The RHS is equal to $$\lim_{n\to\infty}\prod_{k=1}^{3n}\left(1-x^k\right).$$ Do you now see why they’re equal?
A: Assuming that $x$ is a value so that $\prod_{n=1}^{\infty} (1-x^n)$ converges and is meaningful then:
$\prod_{n=1}^{\infty} (1-x^n)=$
$(1-x^1)(1-x^2)(1 -x^3)(1-x^4).......... = $
$[(1-x^3)(1-x^2)(1-x^1)][(1-x^6)(1-x^5)(1-x^4)][(1-x^9)(1-x^8)(1-x^7)].......$
$\prod_{n=1}^{\infty}[(1-x^{3n})(1-x^{3n-1})(1-x^{3n-2})]$
That's probably not the most sophisticated argument...
A much more sophisticated one, and probably closer to what Apostol had in mind is
Every natural number can be uniquely written as $3n$ or $3n-1$ or $3n-2$ and a specific natural $n$, and for any natural $n$ you can determine a unique trio of such natural numbers.
So if $A_k = \{3k, 3k-1, 3k-2\}$ then we know
$\mathbb N = \sum_{k=1}^{\infty} A_k$.
And as The $A_k$ are disjoint we know:
$\prod_{n=1}^{\infty} a_n =$
$\prod_{n\in \mathbb N} a_n = $
$\prod_{n\in \cup_{k=1}^\infty A_k} a_n = $
$\prod_{k=1}^{\infty} [\prod_{n\in A_k} a_n] =$
$\prod_{k=1}^{\infty}[a_{3k}\cdot a_{3k-1} \dot a_{3k-2}]$
