A question about generating function related to weakly decreasing function

Question

Let $f:[n]\to \mathbb{N}$, then there is a formula $\sum_{f(1)\ge f(2)\ge\cdots\ge f(n)\ge 0}q^{f(1)+f(2)+\cdots+f(n)}=\frac{1}{(1-q)(1-q^2)\cdots(1-q^n)}$
For the case $n=1$, since there is a single $f$ s.t. $f(1)=k\ge 0$, by $\frac{1}{1-q}=1+q+\cdots+q^k+\cdots$, it's true. But how to view it when $n$ is larger than 1?

Answer 1

I find $q^{f(2)+f(1)}=q^{f(2)}q^{f(1)}=q^{f(2)-f(1)}q^{2f(1)}$, since $f(2)-f(1)\ge 0$ and has no other constrant, therefore $\sum_{f(1)\ge f(2)\ge 0}q^{f(2)-f(1)}q^{2f(1)}=\frac{1}{1-q}\frac{1}{1-q^2}$
The same method works for other $n$.