Coefficients of polynomial $f_n(q)=\prod_{k=1}^{n}(1+q^k)$

Too long for a comments.

Openning the brackets, you get $q^k$ exactly when you can write $$k=i_1+i_2+....+i_j$$ for $1\leq i_1< i_2< .... < u_j \leq n$

Therefore $c_n(k)=$ is the number of ways $k$ can be written as a sum of distinct integers in $\{1,2,3,.., n\}$.

Alternately, if for each $ A \subset \{ 1,2,3,..,n \}$ you define $f(A)=\sum_{k \in A} k$, then $c_n(k)$ is simply the number of subset $A \subset \{ 1,2,3,..,n \}$ for each $f(A)=k$.

Now, the relation $$c_n(k)=c_n(n(n+1)/2-k) \,.$$ becomes obvious: If $f(A)=k$ then $f(\{ 1,2,..., n\} \backslash A)= n(n+1)/2-k$. In other words, the complement is a bijection between the number of subset $A \subset \{ 1,2,3,..,n \}$ such that $f(A)=k$ and the number of subset $A \subset \{ 1,2,3,..,n \}$ such that $f(A)=n(n+1)/2-k$.

You can also see $$\sum_{k=0}^{n}c_n(k)=2^n.$$ as a consequence that there are exatly $2^n$ subsets on $\{1,2,..,n\}$.