Series to Sum Equation I have this series and I want to convert it to the Summation form.
$1+(1+\alpha _{1})+(1+\alpha _{1})(1+\alpha _{2})+(1+\alpha _{1})(1+\alpha _{2})(1+\alpha _{3})+...$
I know that the next series is the result of multiplying by the previous series.
Let the $a _{1} = 1$, and $a _{2} = a _{1}(1+\alpha _{1})$. This is what I mean. But, then how to change all of those series to the Summation form?
 A: We can write finite products using the product symbol $\prod$ as
\begin{align*}
\left(1+\alpha_1\right)\left(1+\alpha_2\right)\cdots \left(1+\alpha_n\right)=\prod_{k=1}^n\left(1+\alpha_k\right)
\end{align*}
With this notation we can write the series as
\begin{align*}
&1+\left(1+\alpha_1\right)+\left(1+\alpha_1\right)\left(1+\alpha_2\right)+\cdots\\
&\qquad=1+\prod_{k=1}^1\left(1+\alpha_k\right)+\prod_{k=1}^2\left(1+\alpha_k\right)+\cdots\\
&\qquad=1+\sum_{n=1}^\infty\prod_{k=1}^n\left(1+\alpha_k\right)\\
&\qquad\,\,\color{blue}{=\sum_{n=0}^{\infty}\prod_{k=1}^n\left(1+\alpha_k\right)}\tag{1}
\end{align*}
In the last line we note the empty product $\prod_{k=1}^0\left(1+\alpha_k\right)$ with upper bound less than lower bound is by definition $1$.
If we like we can also expand the products. Let $[n]=\{1,2,\ldots,n\}$ denote the $n$ element set of natural numbers $1$ to $n$. We can write
\begin{align*}
\prod_{k=1}^n\left(1+\alpha_k\right)=\sum_{S\subset[n]}\prod_{j\in S}\alpha_j\tag{2}
\end{align*}
An example of (2) with small $n=2$: We obtain
\begin{align*}
\prod_{k=1}^2\left(1+\alpha_k\right)&=\left(1+\alpha_1\right)\left(1+\alpha_2\right)\\
&=1+\alpha_1+\alpha_2+\alpha_1\alpha_2\tag{3}
\end{align*}
and the four terms in (3) correspond to the subsets $\emptyset,\{1\},\{2\},\{1,2\}=[2]$ as stated for general $[n]$ in the sum of (2).

Combining (1) and (2) we can write
\begin{align*}
&1+\left(1+\alpha_1\right)+\left(1+\alpha_1\right)\left(1+\alpha_2\right)+\cdots\\
&\qquad=\sum_{n=0}^{\infty}\prod_{k=1}^n\left(1+\alpha_k\right)\\
&\qquad \color{blue}{=\sum_{n=0}^{\infty}\sum_{S\subset[n]}\prod_{j\in S}\alpha_j}
\end{align*}

