Suppose that we have a commutative, associative ring $R$ which we use to generate the polynomial $R[x]$. Then $p(x) \in R[x]$ is of the form $p(x) = \sum_{i=0}^n a_i x^i$ for some $n \in \mathbb{N}$.
Now, I wish to understand why $R[x]$ is itself a commutative, associative ring. I am happy showing all of the necessary properties to show this to be the case, except for associativity of multiplication.
I understand that given $p(x), q(x) \in R[x]$, such that
$p(x) = \sum_{i=0}^n a_i x^i, \ \ q(x) = \sum_{i=0}^m b_i x^i$
for some $m, n \in \mathbb{N}$, where $a_i, b_i \in R, \forall i$, we may (quite naturally) define their product
$p(x)q(x)= \left(pq \right)(x):=\sum_{k=0}^{n+m} \sum_{i+j=k, \ i,j \in \mathbb{N}_0} a_i b_j x^k.$
However, I get rather lost when I introduce another polynomial, $r(x) \in R[x]$ such that
$r(x) = \sum_{i=0}^o c_i x^i$
for some $o \in \mathbb{N}$, where $c_i \in R, \forall i$ and attempt to show that
$\left(p(x) q(x) \right) r(x) = p(x) \left(q(x) r(x) \right).$
Would any kindly soul be able to show me how?