How to show this identity $\prod_{q=1}^k\frac{1}{1-qz}=\sum_{j=1}^{k}jz\prod_{q=1}^j\frac{1}{1-qz}+1$ avoiding a proof by induction 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.
 A: 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

*

*$\prod\limits_{q=1}^0(\cdots)$ should be interpreted as an empty product and always evaluate to $1$.

A: As commented above, if you expand the LHS you get $$\prod _{q=1}^k \left (\sum _{i\geq 0}{q^iz^i}\right )=\sum _{n\geq 0}\left (\sum _{a_1+a_2+\cdots +a_k =n}1^{a_1}2^{a_2}\cdots k^{a_k}\right )z^n,$$
let $$j=\max _{a_{\ell }\neq 0}{\ell},$$
then $a_{\ell}\geq 1$ and so you can filter this sum as follows
$$
\begin{align*}
\sum _{n\geq 1}\left (\sum _{a_1+a_2+\cdots +a_k =n}1^{a_1}2^{a_2}\cdots k^{a_k}\right )z^n&=\sum _{n\geq 1}\left (\sum _{j=1}^k\sum _{\substack{a_1+a_2+\cdots +a_j =n\\a_j>0}}1^{a_1}2^{a_2}\cdots j^{a_j}\right )z^n\\ &=\sum _{n\geq 1}\left (\sum _{j=1}^kj\sum _{a_1+a_2+\cdots +a_j-1 =n-1}1^{a_1}2^{a_2}\cdots j^{a_j-1}\right )z^n\\
&=\sum _{j=1}^kjz\sum _{n\geq 1}\left (\sum _{a_1+a_2+\cdots +a_j-1 =n-1}1^{a_1}2^{a_2}\cdots j^{a_j-1}\right )z^{n-1}
\end{align*}
$$
Using the same expansion you get the result.
