Proving that the multiplication on a formal derivation ring is associative Let $R$ be a ring , $\delta$ a derivation of $R$, and $x$ an indeterminate. Let $R[x,\delta]$ be for the formal differential operator ring over $R$ (so for $a \in R$, $xa = ax + \delta(a)$). Multiplication on $R[x,\delta]$ is defined by
\begin{align*}
\left(\sum_i a_ix^i\right)\left(\sum_j b_jx^j\right) = \sum_k \left( \sum_{l=0}^k \sum_{i\ge l} {i \choose l} a_i \delta^{i-l}(b_{k-l}) \right) x^k.
\end{align*}
I want to show that this operation is associative. My work is as follows (this will get messy...):
\begin{align}
 &\left [\left(\sum_i a_ix^i\right) \left(\sum_j b_jx^j\right)\right] \left(\sum_k c_kx^k\right) 
\\ & = \left (\sum_m \left( \sum_{l=0}^m \sum_{i\ge l} {i \choose l} a_i \delta^{i-l}(b_{m-l}) \right) x^m \right) \left(\sum_k c_kx^k\right)
\\ & = \sum_n \left ( \sum_{r=0}^n \sum_{s \ge r} {s \choose r} \left( \sum_{l=0}^s \sum_{i\ge l} {i \choose l} a_i \delta^{i-l}(b_{s-l}) \right) \delta^{s-r}(c_{n-r}) \right ) x^n \\
& = \sum_n \left ( \sum_{r=0}^n \sum_{s \ge r} \sum_{l=0}^s \sum_{i\ge l} {s \choose r} {i \choose l} a_i \delta^{i-l}(b_{s-l}) \delta^{s-r}(c_{n-r}) \right ) x^n.
\end{align}
On the other hand, we have (and note that my choice of summation indices aren't great...):
\begin{align*}
 &\left(\sum_i a_ix^i\right) \left [ \left(\sum_j b_jx^j\right) \left(\sum_k c_kx^k\right) \right] \\
& = \left(\sum_i a_ix^i\right) \left (\sum_m \left( \sum_{l=0}^m \sum_{k\ge l} {k \choose l} b_k \delta^{k-l}(c_{m-l}) \right) x^m \right) \\
& = \sum_n \left( \sum_{r=0}^n \sum_{s \ge r} {s \choose r} a_s \delta^{s-r} \left( \sum_{l=0}^{n-r} \sum_{k \ge l} {k \choose l}b_k \delta^{k-l}(c_{n-r-l}) \right)\right ) x^n \\
& = \sum_n \left( \sum_{r=0}^n \sum_{s \ge r} \sum_{l=0}^{n-r} \sum_{k \ge l} {s \choose r} {k \choose l} a_s \delta^{s-r} \left(b_k \delta^{k-l}(c_{n-r-l}) \right)\right ) x^n \\
& = \sum_n \left( \sum_{r=0}^n \sum_{s \ge r} \sum_{l=0}^{n-r} \sum_{k \ge l} \sum_{t=0}^{s-r}{s \choose r} {k \choose l} {s-r \choose t}a_s \delta^{s-r-t}(b_k) \delta^{t+k-l}(c_{n-r-l})\right ) x^n.
\end{align*}
That's pretty ugly. I now want to simplify the right hand side and relabel my indices to obtain the earlier ugly expression. Since I have an extra summation, I think I need to use some binomial coefficient identity, but can't seem to find one that fits this problem. Can anyone provide help?

I've sorted this out now, so I'll post my solution.
It's sufficient to consider products of monomials. We have
\begin{align*}
\Big((ax^r)(bx^s)\Big)(cx^t) &= \left(a \left(\sum_{i=0}^r {r \choose i} \delta^{r-i}(b)x^i \right) x^s\right) cx^t \\
&= \sum_{i=0}^r {r \choose i} a \delta^{r-i}(b) x^i \Big (x^{s} c \Big) x^t \\
&= \sum_{i=0}^r {r \choose i} a \delta^{r-i}(b) x^i \left (\sum_{j=0}^s {s \choose j} \delta^{s-j}(c) x^j \right) x^t \\
&= \sum_{i=0}^r \sum_{j=0}^s {r \choose i}{s \choose j} a \delta^{r-i}(b) \Big( x^i  \delta^{s-j}(c) \Big) x^{j+t} \\
&= \sum_{i=0}^r \sum_{j=0}^s {r \choose i}{s \choose j} a \delta^{r-i}(b) \left(\sum_{k=0}^i {i \choose k} \delta^{i-k+s-j}(c) x^k \right) x^{j+t} \\
&= \sum_{i=0}^r \sum_{j=0}^s \sum_{k=0}^i {r \choose i}{s \choose j} {i \choose k} a \delta^{r-i}(b) \delta^{i-k+s-j}(c) x^{k+j+t}\,.
\end{align*}
On the other hand,
\begin{align*}
(ax^r)\Big((bx^s)(cx^t)\Big) &= (ax^r)\left(b \left(\sum_{j=0}^s {s \choose j} \delta^{s-j}(c)x^j \right) x^t \right) \\
&= \sum_{j=0}^s {s \choose j} a \Big(x^r b \Big) \delta^{s-j}(c)x^{j+t} \\
&= \sum_{j=0}^s {s \choose j} a \left( \sum_{i=0}^r {r \choose i} \delta^{r-i}(b) x^i \right) \delta^{s-j}(c)x^{j+t} \\
&= \sum_{i=0}^r \sum_{j=0}^s {r \choose i} {s \choose j} a  \delta^{r-i}(b) \Big( x^i \delta^{s-j}(c) \Big) x^{j+t} \\
&= \sum_{i=0}^r \sum_{j=0}^s {r \choose i} {s \choose j} a  \delta^{r-i}(b) \left( \sum_{k=0}^i {i \choose k} \delta^{i-k+s-j}(c) x^k \right) x^{j+t} \\
&= \sum_{i=0}^r \sum_{j=0}^s \sum_{k=0}^i {r \choose i}{s \choose j} {i \choose k} a \delta^{r-i}(b) \delta^{i-k+s-j}(c) x^{k+j+t}\,,
\end{align*}
and so
\begin{align*}
\Big((ax^r)(bx^s)\Big)(cx^t) = (ax^r)\Big((bx^s)(cx^t)\Big)
\end{align*}
as required.
