Fadeev and Kashaev define the quantum dilogarithm by $$ \Psi(x) = \prod_{n=1}^\infty (1 - x q^n) $$ for $|q| < 1$. For $q = \exp(\epsilon)$, $\Re \epsilon < 0$, they say the asymptotic expansion $$ \Psi(x) = \frac{1}{\sqrt{1 - x}} \exp( \operatorname{Li}_2(x) / \epsilon)(1 + O(\epsilon)) $$ as $\epsilon \to 0$ is "easy to see," but I'm having trouble deriving it.
Here $\operatorname{Li}_2$ is the dilogarithm $$ \operatorname{Li}_2(x) = - \int_0^x \frac{\log( 1-t )}{t} \, dt = \sum_{n=1}^\infty \frac{x^n}{n^2}. $$ I've tried expanding $$ \log \left( \Psi(x) \sqrt{1-x} \exp(- \epsilon^{-1} \operatorname{Li}_2(x)) \right) = \log \Psi(x) + \frac{1}{2} \log(1 - x) - \frac{1}{\epsilon} \operatorname{Li}_2(x). $$ but this doesn't seem like it can work: if you expand $\log \Psi(x)$ in $\epsilon$ there's only positive powers, so I don't see how you can get a cancellation with the $\epsilon^{-1} \operatorname{Li}_2(x)$.
Not exactly an answer, but this may help.
Define the coefficients $a_n(q)$ by $$f_q(x)=\prod_{m\ge1}(1-xq^m)=\sum_{n\ge0}a_n(q)x^n.$$ Since $f_q(0)=1$, we have $a_0(q)=1$. Then note that $$\begin{align} \prod_{m\ge1}(1-xq^m)&=(1-xq)\prod_{m\ge2}(1-xq^m)\\ &=(1-xq)\prod_{m\ge1}(1-xq^{m+1})\\ &=(1-xq)\prod_{m\ge1}(1-(xq)q^{m})\\ &=(1-xq)f_q(xq). \end{align}$$ Thus $$\begin{align} \sum_{n\ge0}a_n(q)x^n&=(1-xq)\sum_{n\ge0}a_n(q)q^nx^n\\ &=\sum_{n\ge0}a_n(q)q^nx^n-\sum_{n\ge0}a_n(q)q^{n+1}x^{n+1}\\ &=a_0(q)(qx)^0+\sum_{n\ge1}a_n(q)q^nx^n-\sum_{n\ge1}a_{n-1}(q)q^{n}x^{n}\\ &=1+\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}\\ 1+\sum_{n\ge1}a_n(q)x^n&=1+\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}\\ \sum_{n\ge1}a_n(q)x^n&=\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}. \end{align}$$ Equating coefficients, we have $$a_n(q)=\left(a_n(q)-a_{n-1}(q)\right)q^{n},\qquad n\ge1.$$ This is $$a_n(q)=\frac{q^n}{q^n-1}a_{n-1}(q).$$ And since $a_0(q)=1$, we have $$a_n(q)=\prod_{k=1}^{n}\frac{q^k}{q^k-1}=\frac{(-1)^nq^{n(n+1)/2}}{(q;q)_n},$$ that is $$\prod_{m\ge1}(1-xq^m)=\sum_{n\ge0}\frac{(-1)^nq^{n(n+1)/2}}{(q;q)_n}x^n.$$ Hopefully this can help :)