TITLE: Transform $\prod_{k=0}^{n} (1+x^{2^{k}})$ to $\sum_{k=0}^{n} c_{k}x^{k}$
QUESTION [7 upvotes]: I want to transform the following 
$$\prod_{k=0}^{n} (1+x^{2^{k}})$$
to the canonical form $\sum_{k=0}^{n} c_{k}x^{k}$
This is what I got so far
\begin{align*}
\prod_{k=0}^{n} (1+x^{2^{k}})= \dfrac{x^{2^{n}}-1}{x-1} (x^{2^{n}}+1) \\
\end{align*}
but I don't know how to continue, can someone help me with this?

REPLY [3 votes]: You may also approach this from a combinatorical point fo view:

Expanding would give a sum of $2^{n+1}$ summands.
Each summand is a product of powers of $x$ where the exponent corresponds to a uniquely determined sequence $(b_0,b_1,\ldots , b_n)$ of binary digits $b_i \in {0,1}$ for $i=0,\ldots ,n$ where $0$ means choose $1 = x^0$ and $1$ means choose $x^{2^i}$ from the factor $(1+x^{2^i})$

Hence,
$$\prod_{k=0}^{n} (1+x^{2^{k}}) = \sum_{(b_0,b_1,\ldots , b_n) \in \{0,1\}^{n+1}}x^{\sum_{i=0}^nb_i\cdot 2^i}= \sum_{k=0}^{2^{n+1}-1}x^k$$