Polynomial multiplication is associative

Let $A$ be a ring and $f,g,hin A[X]$. I want to show that $(fg)h=f(gh)$, where $(xy)_n:=sum_{n=j+k}x_jy_k$ for all $ninmathbb{N}$.

Attempt: Let $ninmathbb{N}$. I have to show that $((fg)h)_n=(f(gh))_n$. By definition
$$((fg)h)_n=sum_{n=j+k}(fg)_jh_k=sum_{n=j+k}left(sum_{j=s+t}f_sg_tright)h_k=sum_{n=j+k}sum_{j=s+t}(f_sg_t)h_k.$$
Presumably that last expression is equal to $sum_{n=(s+t)+k}(f_sg_t)h_k$. But why? What is the formal justification?