TITLE: Seeking an elegant proof that these two infinite series are equal
QUESTION [8 upvotes]: Recently I encountered this equality in my work:
\begin{align*}
    &\bigg( 1+\frac{1}{p-1} \bigg)
    \bigg( 1+\frac{1}{p^2-1} \bigg)
    \bigg( 1+\frac{1}{p^3-1} \bigg)
    \cdots\\
    =\quad&
    1+\frac{1}{p-1}\bigg(1+\frac{1}{p-1}\bigg(1+\frac{1}{p^2-1}\bigg(1+\frac{1}{p^2-1}\bigg(1+\frac{1}{p^3-1}\bigg(1+\frac{1}{p^3-1}\bigg(1+\cdots\bigg)
\end{align*}
If $p$ is a real number $\geq 2$, it can be easily checked by showing that the ratio of L.H.S and R.H.S tends to $1$. However, this is not the type of proof I wished for.
I'm seeking for a more elegant proof, that would also work for the case in which $p$ is a complex number. For example, is there any way to find the analytical continuation of it? Also, is there any reference for this equality? I believe there are more of this kind in the literature.

REPLY [2 votes]: This is just a small supplement to the instructive answer from @metamorphy. I couldn't see his argument in the last line at a glance and needed some intermediate steps to fully appreciate his solution. Maybe some others are interested too.
We obtain from the right-hand side of (1)
\begin{align*}
\sum_{n=0}^\infty\frac{q^{n(n-1)}z^n}{(q)_n(z)_n}=\frac{1}{(z)_\infty}\tag{1}
\end{align*}
by multiplying by $1-z$ and taking the limit $z\to 1$:

\begin{align*}
\lim_{z\to 1}\frac{1-z}{(z)_\infty}
&=\lim_{z\to 1}\frac{1-z}{\prod_{n=0}^\infty\left(1-zq^n\right)}\\
&=\lim_{z\to 1}\frac{1}{\prod_{n=1}^\infty\left(1-zq^n\right)}\\
&=\frac{1}{\prod_{n=1}^\infty\left(1-q^n\right)}=\frac{1}{\prod_{n=0}^\infty\left(1-q^{n+1}\right)}\\
&\,\,\color{blue}{=\frac{1}{(q)_{\infty}}}
\end{align*}

and the left-hand side gives

\begin{align*}
\lim_{z\to 1}&(1-z)\sum_{n=0}^\infty\frac{q^{n(n-1)}z^n}{(q)_n(z)_n}\\
&=\lim_{z\to 1}(1-z)\sum_{n=0}^\infty
\frac{q^{n(n-1)}z^n}{\prod_{k=0}^{n-1}\left(1-q^{k+1}\right)\prod_{k=0}^{n-1}\left(1-zq^k\right)}\\
&=\lim_{z\to 1}\sum_{n=0}^\infty
\frac{q^{n(n-1)}z^n}{\prod_{k=0}^{n-1}\left(1-q^{k+1}\right)\prod_{k=1}^{n-1}\left(1-zq^k\right)}\\
&=\sum_{n=0}^\infty
\frac{q^{n(n-1)}z^n}{\prod_{k=0}^{n-1}\left(1-q^{k+1}\right)\prod_{k=1}^{n-1}\left(1-q^k\right)}\\
&=\sum_{n=0}^\infty
\frac{q^{n(n-1)}}{\prod_{k=0}^{n-1}\left(1-q^{k+1}\right)\prod_{k=0}^{n-2}\left(1-q^{k+1}\right)}\\
&=\sum_{n=0}^\infty
\frac{\left(1-q^n\right)q^{n(n-1)}}{\prod_{k=0}^{n-1}\left(1-q^{k+1}\right)\prod_{k=0}^{n-1}\left(1-q^{k+1}\right)}\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\frac{\left(1-q^n\right)q^{n(n-1)}}{(q)_{n}^2}}
\end{align*}

showing the validity of OP's identity for $q:=\frac{1}{p}$ with $|q|<1$
\begin{align*}
\color{blue}{\frac{1}{(q)_{\infty}}=\sum_{n=0}^\infty\frac{\left(1-q^n\right)q^{n(n-1)}}{(q)_{n}^2}}
\end{align*}