I am having trouble completing a manipulation connected with distribution transposes. My mathematical objects may be overspecified here, but I prefer to replicate the example exactly.
Let $p$, $q$ be multi-indices of length $n$, $\lambda_p$ a Radon measure, $g$ a $\mathcal{C}^{\infty}(\Omega)$ function and $\phi$ a test function. The text I'm reading claims the following: $$(-1)^{|p|}\sum_{q\le p}{p\choose q}\langle\lambda_p,[(\delta/{\delta x})^{p-q}g](\delta/{\delta x})^q\phi\rangle =\sum_{q\le p}\langle(\delta/{\delta x})^q[(-1)^{|p-q|}\lbrace (\delta/{\delta x})^{p-q}g\rbrace],\phi\rangle$$
I'm unable to get rid of the $^pC_q$ terms on the RHS. Here are my steps:
With $p$, $q$ fixed, $(\delta/{\delta x})^{p-q}g$ is a degree-zero differential operator, which means it is identical to its transpose. This gets me from LHS to $$(-1)^{|p|}\sum_{q\le p}{p\choose q}\langle\{(\delta/{\delta x})^{p-q}g\}\lambda_p,(\delta/{\delta x})^q\phi\rangle.$$
Next, $P:=(\delta/{\delta x})^q$ is a differential operator with constant coefficient. Somewhat confusingly to me, my reference text defines $^tP(\delta/\delta x)=P(-\delta/\delta x)$ in this case. I get by hand that this is $(-1)^{|q|}(\delta/\delta x)^q$. Putting that into the previous expression gives $$(-1)^{|p|}\sum_{q\le p}{p\choose q}\langle(-1)^{|q|}(\delta/{\delta x})^q[\lbrace(\delta/{\delta x})^{p-q}g\rbrace\lambda_p],\phi\rangle.$$
Since $(-1)^{|p|+|q|}=(-1)^{|p+q|}=(-1)^{|p-q|}$, this gives $$\sum_{q\le p}{p\choose q}\langle (-1)^{|p-q|}(\delta/{\delta x})^q[\lbrace(\delta/{\delta x})^{p-q}g\rbrace\lambda_p],\phi\rangle=\sum_{q\le p}{p\choose q}\langle (\delta/{\delta x})^q [(-1)^{|p-q|}\lbrace(\delta/{\delta x})^{p-q}g\rbrace\lambda_p],\phi\rangle$$
After a lot of struggle I remain unable to reach the target expression. A bit of (kind) advice would be appreciated.
P.S. This should be a self-sufficient part of Francois Treves's proof (1995, Theorem 24.4) that every distribution with compact support on an open subset of $\mathbb{R}^n$ is of finite order.