Proof of differential operator identity in DLMF $16.3.5$

Solution 1:

Define the linear operators $\, D(f) \!:=\! \partial_z f, \, Z(f) \!:=\! z f \,$ with $\, DZ(f) \!=\! f \!+\! Z D(f) .\,$ Since the operators $\, Z D\,$ and $\,DZ\,$ differ by the identity operator, they commute. Moreover, we can prove that $\,Z D\,$ and $\,D^n Z^n\,$ commute for all $\,n\,$ by induction.

Using simple algebra and a few steps we can prove that $$ Z DZ(Z^n D^n Z^n) \!=\! Z^{n+1}((n\!+\!1)D^n Z^n \!+\!Z D^{n+1}Z^n). \tag{1} $$ But now, again using simple algebra, $$ Z^{n+1}D^{n+1}Z^{n+1}\!=\! Z^{n+1}((n\!+\!1)D^n Z^n + D^n Z^{n+1}D). \tag{2}$$ Since $\,Z D\,$ and $\,D^n Z^n\,$ commute, the two quantities on the right side of equations $(1)$ and $(2)$ are equal. Thus, $\,(Z DZ)^n = Z^n D^n Z^n$ for all $\,n\,$ by induction.