Is there a combinatorial interpretation for these identities? I recently stumbled across the two seemingly similar identities
$$
\prod_{i\geq 1}\frac{1}{1-xq^i}=\sum_{n\geq 0}\frac{x^nq^n}{(1-q)(1-q^2)\cdots(1-q^n)}
$$
and
$$
\prod_{i\geq 1}(1+xq^i)=\sum_{n\geq 0}\frac{x^nq^{\binom{n+1}{2}}}{(1-q)(1-q^2)\cdots(1-q^n)}.
$$
Out of curiosity, is there some combinatorial interpretation for these identities, to understand why they hold? Thanks.
If this would better be asked as two questions, I am glad to split it into two.
 A: The LHS of the first identity is a generating function
$$\prod_{i \ge 1} \frac{1}{1 - xq^i} = \sum p_{m,n} q^m x^n$$
where $p_{m,n}$ counts the number of ways to partition the number $m$ into a sum of $n$ positive integers. In other words, $p_{m,n}$ counts the number of Ferrers diagrams with $m$ dots and $n$ rows. 
For fixed $n$, given such a Ferrers diagram slice off the leftmost column. The remaining columns form a partition into parts of size at most $n$, and such partitions have generating function $\frac{1}{(1 - q)...(1 - q^n)}$. The leftmost column contributes a factor of $q^n$, and the fact that we started with a Ferrers diagram with $n$ rows contributes a factor of $x^n$. This gives the RHS of the first identity.

The LHS of the second identity counts the number of ways to partition the number $m$ into a sum of $n$ distinct positive integers. To get the RHS, instead of slicing off the leftmost column of the corresponding Ferrers diagram, you can slice off a right triangle with side lengths $n$ and $n$ because the number of dots in each row are distinct (draw a diagram to convince yourself of this). This triangle has ${n+1 \choose 2}$ dots and the rest of the argument is the same as above. 
A: I'll somewhat expound the Qiaochu Yuan's wonderful solution, and show that how it can be translated to algebraic expressions.
Second identity
We start from the second identity
\begin{align}
\prod_{i = 1}^\infty (1 + q^i x)
= \sum_{n = 0}^\infty p^F_{n} x^n.
\tag{1}
\end{align}
Expanding the left hand side and comparing the coefficient of $x^n$, we find
$$
\sum_{1 \le a_1 <  \dots < a_n} q^{a_1 + \cdots + a_n}
=
p^F_n,
\tag{2}
$$
where the sum is carried over all the combinations of $a_1, \dots, a_n$,
satisfying the condition $1 \le a_1 < \dots < a_n$.
Each term in the sum for $p^F_n$ can be represented by a Ferrers diagram,
in which, there are $n$ rows, each with a different number $a_k$ ($1 \le k \le n)$ of dots.  If the rows are labelled from bottom to top, a row with fewer dots lies below a row with more dots.
Now consider the change of variables
\begin{align}
b_k = a_k - a_{k - 1},
\end{align}
with $a_0\equiv 0$, then the condition of $a_k > a_{k-1}$ becomes
$$
b_k \ge 1,
$$
for any $k = 1, \dots, n$.
The inverse relation is
\begin{align}
a_1 &= b_1, \\
a_2 &= a_1 + b_2 = b_1 + b_2, \\
a_3 &= a_2 + b_3 = b_1 + b_2 + b_3, \\
\cdots \\
a_n &= a_{n - 1} + b_n = b_1 + b_2 + \cdots + b_n.
\end{align}
The sum on the exponent of the left hand side of (2) becomes
$$
a_1 + \cdots + a_n
=
n \, b_1 + (n-1) \, b_2 + \cdots + b_n.
$$
and (2) becomes
\begin{align}
p^F_n
&=
\sum_{b_1 \ge 1, \; b_2 \ge 1, \; \cdots, \; b_n \ge 1}
q^{n \, b_1 + (n-1) \, b_2 + \cdots + b_n} \\
&=
\sum_{b_1 \ge 1} q^{n \, b_1}
\sum_{b_2 \ge 1} q^{(n - 1) \, b_2}
\cdots
\sum_{b_n \ge 1} q^{b_n} \\
&=
\frac{q^n}{1-q^n}
\frac{q^{n-1}}{1-q^{n-1}}
\cdots
\frac{q}{1-q} \\
&=
\frac{ q^{n(n+1)/2} }
{(1-q) (1-q^2) \cdots (1-q^n) }.
\end{align}
This is the desired result.
First identity
The case for the first identity is similar 
\begin{align}
\prod_{i = 1}^\infty \left[1 + q^i x + (q^i x)(q^i x) + (q^i x)(q^i x)(q^i x) + \cdots \right]
= \sum_{n = 0}^\infty p^B_{n} x^n.
\tag{3}
\end{align}
Comparing the coefficient of $x^n$, we find
$$
\sum_{1 \le a_1 \le  \dots \le a_n} q^{a_1 + \cdots + a_n}
=
p^B_n,
\tag{4}
$$
Note that “$<$” is replaced by “$\le$” because of the $(q^i x)(q^i x), (q^i x)(q^i x)(q^i x), \dots$ terms.  In terms of the Ferrers diagram, we now allow multiple rows with the same number of dots.  For example, if we choose the term $(q^i x) (q^i x) (q^i x)$ for the $i$th factor, then there are three rows with $i$ dots in the Ferrer diagram.
Then, following the same procedure of changing variables, we now get the condition
$$
b_k \ge 0,
$$
for $k > 1$, and $b_1 \ge 1$.
and
\begin{align}
p^B_n
&=
\sum_{b_1 \ge 1, \; b_2 \ge 0, \; \cdots, \; b_n \ge 0}
q^{n \, b_1 + (n-1) \, b_2 + \cdots + b_n} \\
&=
\sum_{b_1 \ge 1} q^{n \, b_1}
\sum_{b_2 \ge 0} q^{(n - 1) \, b_2}
\cdots
\sum_{b_n \ge 0} q^{b_n} \\
&=
\frac{q^n}{1-q^n}
\frac{1}{1-q^{n-1}}
\cdots
\frac{1}{1-q} \\
&=
\frac{ q^n }
{(1-q) (1-q^2) \cdots (1-q^n) }.
\end{align}
Q.E.D.
