I'm required to write $\prod\limits_{i=1}^n(1-j_{i})$ as a nested sum, where $j_{ll} \neq j_{k}$ if $u \neq k$.
I undestand I'd get something in the form of $1-\sum\limits_{i=1}^n{j_{i}}+...+(-1)^n\prod\limits_{i=1}^nj_{i}$ but I don't know how I could write this as k-nested sums, in the form $\sum^{(k)} (expression)$.
The way I prefer to transform products to sums is: $$\prod_{i=1}^n(1+a_i)=\sum_{I\subseteq [1,n]}\prod_{i\in I}a_i \tag{1}$$ the sum is taken over all subsets of $[1,n]$ for example: $$\prod_{i=1}^2(1+a_i)=\prod_{i\in \emptyset}a_i+\prod_{i\in\{1\}}a_i+\prod_{i\in\{2\}}a_i+\prod_{i\in\{1,2\}}a_i=1+a_1+a_2+a_1a_2$$ And your product can be deduced from the formula $(1)$: $$\prod_{i=1}^n(1-a_i)=\sum_{I\subseteq [1,n]}(-1)^{|I|}\prod_{i\in I}a_i $$ where $|I|$ denotes the cardinal of the set $I$
There is other notation other then $1$ to transform the product to a sum, mainly $$\prod_{i=1}^n(1+a_i)=1+\sum_{1\leq i_1<i_2<\cdots<i_k\leq n}\prod_{i=1}^ka_i \tag{2}$$
where the sum is taken over all ordered tuples $i_1<i_2<\cdots$ and your product will become using this notation: $$\prod_{i=1}^n(1-a_i)=1+\sum_{1\leq k,1\leq i_1<i_2<\cdots<i_k\leq n}(-1)^k\prod_{i=1}^ka_i $$
As an practical example in which there is the two formulas you can see the principale of inclusion-exclusion
