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.

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 $1-e^{\imath\ t}\neq 0$).
