Show that the given products are equal

Question

I'm trying to show that $$\frac{p^{n^2}(p^{n+1}-1)}{p-1}\prod_{j=1}^n(1-p^{-j})=\prod_{j=1}^n (p^{n+1}-p^{j-1})$$

I've started with the RHS and have done the following:

\begin{align*}\prod_{j=1}^n (p^{n+1}-p^{j-1}) & = p^{n(n+1)} \prod_{j=1}^n (1-p^{j-n-2}) \\ & = p^{n(n+1)} \prod_{j=1}^n (1-p^{-j-1}) \end{align*}

What do I need to do next?

Answer 1

You may continue by writing $$ \begin{align*} \prod_{j=1}^n (p^{n+1}-p^{j-1}) & = p^{n(n+1)} \prod_{j=1}^n (1-p^{-j-1}) \\ & = p^{n(n+1)} \prod_{j=2}^{n+1} (1-p^{-j}) \\ & = p^{n(n+1)} \frac{1-p^{-n-1}}{1-p^{-1}}\prod_{j=1}^{n} (1-p^{-j}) \\ & =\frac{p^{n(n+1)}\cdot p \cdot(p^{n+1}-1)}{p^{n+1}(p-1)}\prod_{j=1}^n(1-p^{-j}) \\ & = \frac{p^{n^2}(p^{n+1}-1)}{p-1}\prod_{j=1}^n(1-p^{-j}) \end{align*} $$ as expected.

Answer 2

$$\begin{align}p^{n(n+1)}\prod_{j=1}^n(1-p^{-j-1})&=p^{n^2}p^n\frac{1-p^{-n-1}}{1-p^{-1}}\prod_{j=1}^n(1-p^{-j})\\&=p^{n^2}\frac{p^n-p^{-1}}{1-p^{-1}}\prod_{j=1}^n(1-p^{-j})\\&=p^{n^2}\frac{p^{n+1}-1}{p-1}\prod_{j=1}^n(1-p^{-j})\end{align}$$

First step foiled the $n(n+1)$ and reindexed the product (try expanding a few terms to se what I did).

Second step distributes the $p^n$ to the fraction.

Last step multiplies the fraction by $\frac pp$.