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

Question

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.

Answer 1

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 $1$.

Answer 2

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.