When looking at a nice problem regarding Stirling numbers of the second kind a challenge was to show the validity of
\begin{align*}
\color{blue}{\prod_{q=1}^k\frac{1}{1-qz}=\sum_{j=1}^kjz\prod_{q=1}^j\frac{1}{1-qz}+1\qquad k\geq 1}\tag{1}
\end{align*}
I was keen on finding a proof avoiding induction, since this usually provides additional information about the structure of the identity.
-
One idea was to multiply (1) with $\prod_{q=1}^k(1-qz)$ and consider it as polynomial identity:
\begin{align*}
\prod_{q=1}^k(1-qz)+\sum_{j=1}^kjz\prod_{q=j+1}^k(1-qz)-1=0
\end{align*}
Since the left-hand side is a polynomial of degree $\leq k$ finding $k+1$ pairwise different zeros would prove the identity. -
I also thought the Leibniz product rule in the form
\begin{align*}
\frac{d}{dz}\prod_{q=1}^k(1-qz)=\sum_{j=1}^k(-j)\prod_{{q=1}\atop{q\neq j}}^k(1-qz)
\end{align*}
might be useful.
Regrettably, I was not successful so far. Helpful information how to prove this identity without using induction is much appreciated.
Best Answer
Notice $$jz\prod\limits^j_{q=1}\frac{1}{1 - qz} = (1 - (1 - jz))\prod\limits^j_{q=1}\frac{1}{1 - qz} = \prod\limits^j_{q=1}\frac{1}{1 - qz} - \prod\limits^{j-1}_{q=1}\frac{1}{1 - qz}$$
The sum on LHS is a telescoping sum. As a result,
$$\require{cancel}{\rm LHS} = 1 + \sum_{j=1}^k jz\prod_{q=1}^j\frac{1}{1-qz} = 1 + \left(\prod_{q=1}^k \frac{1}{1-qz} - \color{red}{\cancelto{1}{\color{gray}{ \prod_{q=1}^0\frac{1}{1-qz} }}}\right) = \prod_{q=1}^k\frac{1}{1-qz} $$
Note