I am trying to show that the coefficient of $x^n$, $n\ge 3$ in
$\prod_{k=1}^{\infty} (1+x^{3^k})(1+x^{4^k})(1+x^{5^k})(1+x^{6^k})$
is always positive.
At this point for convenience let $P_n$ be the same product but with an upper value of $k=n$.
I am finding this quite difficult to do using simple methods that don't involve combinatorial reasoning etc (the original problem was combinatorial and there are solutions, but that is not what I am interested in).
I made some progress before hitting a roadblock. What I have done so far:
First I formed the notation $(a,b):=x^a+x^{a+1}+\cdots x^b$ where the exponents are consecutive. Then I noticed that since we are only interested in positivity, we may reduce the coefficients to all $1$'s at any step without hurting the conclusion. I denoted this by $\rightarrow$, eg $(3,6)+(5,9)\rightarrow (3,9)$.
So the problem basically reduces to proving that there exists $(3,\infty)$ in the product's expansion, as this will mean that the coefficient of $x^n$ for all $n\ge 3$ is positive, as required.
At first, it seems like an easy induction problem, because we find that $P_1\rightarrow (3,15)$, and it seems like we can just continue by an easy induction. However the problem is when we reach higher $P_i$'s like $$\frac{P_5}{(1+x^{6^5})}\rightarrow (3,15+5^5+4^5+3^5+6^5+5^5+\cdots +6^2+5^2+4^2+3^2).$$
(We will just denote the long expression in the bracket from now on by $[5^5]$.
Then we find that the next step, ie multiplying by $1+x^{6^5}$, leads to $\rightarrow (3,[5^5])+(6^5+3,[6^5])$, where unlike in previous cases, we now don't have $6^5+3<[5,5]$ and thus cannot "join" the two brackets together by the $\rightarrow $ operation. This ends up not being a problem a little further along, and we end up again with a joined bracket of the form $(3,x)$ for large $x$. However, in general I have not found a way of dealing with such cases where $6^k$ comes up, and it seems to get more and more complicated.
Is there a nice approach that uses similar ideas (ie not a completely different solution that uses combinatorial interpretations of the generating function etc)?