Computing the product $(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n$ I want to compute the product
$$
(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n,
$$
for a natural number $n$. For $n$ equal to 0 or 1, the computation is very simple but for such a low number as 2 the brute force calculation begins to be rather cumbersome and I cannot see any pattern emerging. I tried to find some connection with the Rodrigues' formula for the Hermite polynomials but I could not. 
These operators come up in the algebraic approach to the quantum harmonic oscillator.
Explicit Example
To avoid any misunderstanding, I am going to show explicitly the computation for the case $n=1$:
$$
(\frac{d}{dx}+x)(-\frac{d}{dx}+x)=-\frac{d^2}{dx^2}+1+x\frac{d}{dx}-x\frac{d}{dx}+x^2=-\frac{d^2}{dx^2}+x^2+1.
$$
One can think of a function $f$ the operators are acting on. For example,
$$
(\frac{d}{dx}\circ x) f= (\frac{d}{dx}x)f+x\frac{d}{dx}f=(1+\frac{d}{dx})f,  
$$
then
$$
\frac{d}{dx}\circ x=1+x\frac{d}{dx}
$$

For convenience let us rewrite $x,\partial_x$ with $a,b$ so $[a,b]=x\partial_x-\partial_x x=-1$ (in the operator sense on the Schwartz space).

Lemma. For all $n\in\mathbb N$
$$
[(a+b)^n,a-b]=2n(a+b)^{n-1}
$$

Proof. As darij pointed out, one has $[a+b,a-b]=2$ (i.e. the case $n=1$). The trick then is
$$
\begin{split}
[(a+b)^{n+1},a-b]&=(a+b)[(a+b)^n,a-b]+[a+b,a-b](a+b)^n\\
&=(a+b)2n(a+b)^{n-1}+2(a+b)^{n}=2(n+1)(a+b)^{(n+1)-1}
\end{split}
$$
which concludes the proof via induction. $\square$

Proposition. For all $n\in\mathbb N_0$
$$
(a+b)^n(a-b)^n=\prod_{j=1}^n (a^2-b^2+(2j-1))
$$

Proof. ($n=0$ is obvious). Note that $(a-b)(a+b)=a^2-b^2-1$. Using the previous lemma
$$
\begin{split}
(a+b)^{n+1}(a-b)^{n+1}&=[(a+b)^{n+1},a-b](a-b)^n+(a-b)(a+b)(a+b)^n(a-b)^n\\
&=2(n+1)(a+b)^n(a-b)^n +(a^2-b^2-1)(a+b)^n(a-b)^n\\
&=\big( a^2-b^2+2n+1)(a+b)^n(a-b)^n=\prod_{j=1}^{n+1} (a^2-b^2+(2j-1))
\end{split}
$$
which again concludes the proof via induction. $\square$
This result reproduces the cases (aside from $n=0$, obvious)


*

*$(a+b)(a-b)=a^2-b^2+1$

*Making use of $[a^2,b^2]=-4ab-2$ (similar techniques) one gets
$$
\begin{split}
(a+b)^2(a-b)^2=(a^2-b^2+1)(a^2-b^2+3)&=a^4-a^2b^2-b^2a^2+4a^2-4b^2+b^4+3\\
&=a^4-2a^2b^2-4ab+4a^2-4b^2+b^4+1
\end{split}
$$
etc... I feel like this formula is the best thing one can hope for in terms of structure.
Edit: Thanks darij for the +200 rep!
