Complex number: equation I would like an hint to solve this equation: $\forall n\geq 1$
$$\sum_{k=0}^{2^n-1}e^{itk}=\prod_{k=1}^{n}\{1+e^{it2^{k-1}}\} \qquad \forall t \in \mathbb{R}.$$
I went for induction but without to much success; I will keep trying, but if you have an hint...
Many thanks.
 A: Hint: Note that the sum on the left is a geometric series. Without worrying yet about division by $0$, we get that the sum is 
$$\frac{1-e^{2^nit}}{1-e^{it}}.$$
For the induction, perhaps use the fact that
$$1-e^{2^{n+1}it}=(1-e^{2^{n}it})(1+e^{2^{n}it}).$$
Once you have worked through the idea, note the following slicker but equivalent calculation. There are two cases, $e^{it}=1$ and $e^{it}\ne 1$. In the second case, multiply each side by $1-e^{it}$. Admire the beautiful collapse on both the left-hand side and the right-hand side. 
For the right-hand side, which is less familiar, look for example at 
$$(1+a)(1+a^2)(1+a^4)(1+a^8),$$
and multiply on the left by $1-a$. 
A: Fix $n\in \mathbb{N}$ and $t\in \mathbb{R}$.


*

*If $t=0 \mod 2\pi$, your equality is obviously true, for it reduces to:
$$\sum_{k=0}^{2^n -1} 1 =2^n = \prod_{k=1}^n 2$$
(remember that $e^{\imath\ t}$ is $2\pi$-periodic).

*Now, assume $t\neq 0 \mod 2\pi$. Evaluate separately:
$$\begin{split}
(1-e^{\imath\ t})\ \sum_{k=0}^{2^n-1}e^{itk} &= \sum_{k=0}^{2^n-1}e^{itk} - \sum_{k=1}^{2^n}e^{itk}\\
&= 1-e^{\imath\ t\ 2^n}
\end{split}$$
and:
$$\begin{split}
(1-e^{\imath\ t})\ \prod_{k=1}^{n} (1+e^{it2^{k-1}}) &= (1-e^{\imath\ t})\ (1+e^{\imath\ t})\ \prod_{k=2}^{n} (1+e^{it2^{k-1}})\\
&= (1-e^{\imath\ t\ 2})\ (1+e^{\imath\ t\ 2})\ \prod_{k=3}^{n} (1+e^{it2^{k-1}})\\
&= (1-e^{\imath\ t\ 4})\ (1+e^{\imath\ t\ 4})\ \prod_{k=4}^{n} (1+e^{it2^{k-1}})\\
&= \cdots\\
&= (1-e^{\imath\ t\ 2^{n-1}})\ (1+e^{\imath\ t\ 2^{n-1}})\\
&= 1-e^{\imath\ t\ 2^n}\; ;
\end{split}$$
therefore you got:
$$(1-e^{\imath\ t})\ \sum_{k=0}^{2^n-1}e^{itk} = (1-e^{\imath\ t})\ \prod_{k=1}^{n} (1+e^{it2^{k-1}})\; ,$$
which yields the desidered equality (because $1-e^{\imath\ t}\neq 0$).
A: Let us consider the polynomial $P = \sum_{k=0}^{2^n -1 } X^k $ and $P_2 = \prod_{k=1}^{n} (1+X^{2^{k-1}})$. There exist  a full combinatorial proff of this result. Hint : if you develop the second term and examine all power of $X$ you could find all the binary development of integer between $0$ and $2^{n}-1$. If you want that I complete the proof, ask me. 
