With $e_i\sim\exp(1)$ why does $\prod_{j=1}^n\exp(ite_j) = (\frac{1}{1-it})^n$? We have $S_n=\sum_{i=1}^ne_i$ with $e_i\sim\exp(1)$ why does $\prod_{j=1}^n\exp(ite_j) = \left(\frac{1}{1-it}\right)^n$? I just want to understand the following line from my notes and hope it is correct: $$\mathbb{E}e^{itS_n} = \mathbb{E}\exp\left({it\sum_{j=1}^ne_j}\right) = \prod_{j=1}^n\mathbb{E}\exp(ite_j) = \left(\frac{1}{1-it}\right)^n$$ Is it related to the definition as in $$\exp(x) = \lim_{n\to\infty}\left(1 + \frac{x}{n}\right)^{n}$$ I have tried looking at the product of terms $\lim_{n\to\infty}\left(1+\frac{ite_j}{n}\right)^n$ but got nowhere. Which made me think it is to do with the $e_j$ being exponentially distributed so $$\left(1+\frac{ite^{-x}}{n}\right)^n$$ And I can't see how to proceed. I suspect I am missing something simple.

Since $e_j\sim Exp(1)$, the density of $e_j$ is $f\left( x \right) = e^{-x}$ if $x\ge 0$ and $f(x)=0$ if $x<0$. Then $$E\left( {{e^{it{e_j}}}} \right) = \int\limits_{ - \infty }^\infty {{e^{itx}}f\left( x \right)dx} = \int\limits_0^\infty {{e^{itx}}{e^{ - x}}dx} = \int\limits_0^\infty {{e^{\left( {it - 1} \right)x}}dx} = \frac{1} {{1 - it}}.$$