TITLE: what else is in $\prod_{j=1}^n(1+q^j)$?
QUESTION [25 upvotes]: From time to time, I run into the finite product $\prod_{j=1}^n(1+q^j)$. And, the more it happens, the more fascinated I've become. So, herein, I wish to get help in collecting such results. To give some perspective into what I look for, check out the below examples. First, some nomenclature: $(q)_k=(1-q)(1-q^2)\cdots(1-q^k)$ and $\binom{n}k_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$.
(0) It's almost silly, but the sum of elementary functions of the specialization $\pmb{q}=(q,q^2,\dots,q^n)$:
$$e_0(\pmb{q})+e_1(\pmb{q})+\cdots+e_n(\pmb{q})=\prod_{j=1}^n(1+q^j).$$
(1) The classical $q$-binomial theorem, which results from counting restricted distinct partitions or number of weighted tilings:
$$\sum_{k=0}^nq^{\binom{k+1}2}\binom{n}k_q=\prod_{j=1}^n(1+q^j).$$
(2) I can't remember where I saw this (do you?) but 
$$\sum_{k=0}^nq^k\binom{n}k_{q^2}=\prod_{j=1}^n(1+q^j).$$
(3) The $H$-polynomial of a symplectic monoid $MSp_n$ (see this paper, page 13): 
$$\sum_{k=0}^n(-1)^kq^{k^2}\binom{n}k_{q^2}^2\prod_{i=1}^k(1-q^{2i})\prod_{j=1}^{n-k}(1+q^j)^2=\prod_{j=1}^{2n}(1+q^j),$$
although the authors did not seem to be aware of the RHS.

QUESTION. Can you provide such formulas (in any field) with the same RHS (always a finite product) as in above, together with resources or references? 

Thank you.

REPLY [2 votes]: Let $exp(z,q)=\sum_{k=0}^{\infty}z^k/[k]_q!$ be the usual  $q-$analogue of the exponential function $e^x.$
The identity
$\sum_{k=0}^n{q^k}{\binom{n}{k}_{q^2}}=\prod_{j=1}^n(1+q^j)$ 
can be obtained by comparing coefficients in 
$$exp(\frac{z}{1+q},q^2)exp(\frac{qz}{1+q},q^2)=exp(z,q),$$
which is a natural $q-$analogue of $e^\frac{x}{2} e^\frac{x}{2}=e^x.$
This identity occurs in Séminaire Lotharingien de Combinatoire, B05a (1981), but is perhaps older.