Sequence from a generating function? 
Find the sequence $A_0, A_1, A_2, \ldots$ if its generating function $A(x) = A_0 + A_1 x + A_2 x^2 + \cdots$ is $$A(x) = \prod_{n=1}^{\infty}(1-q^nx) .$$

Well, i need to find the expansion of $A(x)$ in terms of powers of $x$. For that I take the log of both the sides to get 
$$\log(A(x)) = \sum_{n=1}^{\infty}\log(1-q^nx)$$
Now how to proceed further? Or is there any other method to expand $A(x)$.
 A: We have $$A(x)=\prod_{n \geq 1} (1-xq^n)=\sum_{n \geq 0} \frac{(-1)^nx^nq^{\frac{n(n+1)}{2}}}{(q;q)_n},$$ where $(q;q)_n := (1-q) \cdots (1-q^n).$
Proof with Recursion:  Note that $A(x)= (1-xq)A(xq).$  Thus, if $A(x)= \sum_{n \geq 0} A_n x^n$, then \begin{align*} \sum_{n \geq 0} A_nx^n &= \sum_{n \geq 0} A_nx^nq^n - \sum_{n \geq 0} A_nx^{n+1}q^{n+1} \\ &= 1 + \sum_{n \geq 1} x^n(q^nA_n - A_{n-1}q^{n}),\end{align*} since $A_0=1$.  Thus, $$A_n = q^nA_n - A_{n-1}q^{n} \implies A_n = \frac{-q^{n}A_{n-1}}{1-q^n}.$$  You can show that this recursion is also satisfied by the coefficients of the RHS, which concludes the proof.
Combinatorial Proof  Note that $$A(x) = \sum_{\lambda \in \mathcal{D}} (-x)^{\ell(\lambda)}q^{|\lambda|},$$ where $\mathcal{D}$ is the set of partitions $\lambda$ into distinct parts, $\ell(\lambda)$ is the number of parts and $|\lambda|$ is the size of the partition.  Thus, if we can show that the RHS is also this generating function, then the two are equal.
Namely, observe that in each distinct parts partition $\lambda$ there is a unique triangle as below $$\begin{matrix} {\color{red} \bullet} & {\color{red} \bullet} & {\color{red} \bullet} & \bullet & \bullet & \bullet \\ {\color{red} \bullet} & {\color{red} \bullet} & \bullet & & & \\ {\color{red} \bullet} & & & & & \end{matrix}.$$  Here the red dots sum to $1+2+3=\frac{3(4)}{2}$ which is counted by $(-x)^{n}q^{\frac{n(n+1)}{2}}$ and the black dots form an arbitrary partition counted by $\frac{1}{(q;q)_n}$.  Summing over $n=\ell(\lambda)$ gives the claim.
$q$-Binomial Theorem  Cauchy's $q$-deformation of the binomial theorem states $$\sum_{n \geq 0} \frac{(a;q)_n}{(q;q)_n} z^n = \frac{(az;q)_{\infty}}{(z;q)_{\infty}},$$ which is easy to verify through recursion.  For your result, one may substitute $z \to -xqa^{-1}$ and then take $a \to \infty.$
