Alternative proof of $\prod_{n=1}^{+\infty}\frac{e^{it/2^n}+1}{2}=\frac{e^{it}-1}{it}$

Question

Let $t\in \mathbb{R}$. I want an alternative proof of the following identity $$\prod_{n=1}^{+\infty}\frac{e^{it/2^n}+1}{2}=\frac{e^{it}-1}{it} \quad(\star)$$ I've came up with this identity observing that if $X_1, X_2, \ldots$ are independent random variables with Bernoulli distribution of parameter $1/2$, then the variable $$Y=\sum_{n=1}^{+\infty}\frac{X_n}{2^n}$$ has uniform distribution over $[0, 1]$. Comparing the characteristic functions of $Y$ and the uniform distribution we deduce $(\star)$.

Answer 1

$$\prod\limits_{n=1}^{+\infty}\dfrac{e^{it/2^n}+1}{2}=\lim\limits_{N\to\infty}\prod\limits_{n=1}^N\dfrac{e^{it/2^n}+1}{2}=\lim\limits_{N\to\infty}\dfrac{(e^{it/2}+1)\cdots(e^{it/2^{N-1}}+1)(e^{it/2^N}+1)}{2^N}=\dots$$

Let's multiply and divide the expression by $(e^{it/2^N}-1)$ and then use identity $(a-b)(a+b)=a^2-b^2$:

$$\dots=\lim\limits_{N\to\infty}\dfrac{(e^{it/2}+1)\cdots(e^{it/2^{N-1}}+1)(e^{it/2^N}+1)(e^{it/2^N}-1)}{2^N(e^{it/2^N}-1)}=\lim\limits_{N\to\infty}\dfrac{(e^{it/2}+1)\cdots(e^{it/2^{N-1}}+1)(e^{it/2^{N-1}}-1)}{2^N(e^{it/2^N}-1)}=\dots=\lim\limits_{N\to\infty}\dfrac{e^{it}-1}{2^N(e^{it/2^N}-1)}$$

In denominator we have $$\lim\limits_{N\to\infty}2^N(e^{it/2^N}-1)=it\lim\limits_{N\to\infty}\frac{e^{it/2^N}-1}{it/2^N}=it$$ so finally $$\prod\limits_{n=1}^{+\infty}\dfrac{e^{it/2^n}+1}{2}=\dfrac{e^{it}-1}{it}$$