$X_1, X_2, .....X_n$ is a colletion of Random Variables. They are said to be $\textit{multiplicative system}$ if, for any $1 \leq k \leq n$ and for any set of $k$ indices $1 \leq i_1 < i_2 ....i_k \leq n$ -
$$E[X_{i_1}X_{i_2}.....X_{i_k}] = 0 $$ I need to prove that if $X_1, X_2, ....X_n$ are multiplicative system then, $$E\bigg[ \prod_{i=1}^n(a_iX_i + b_i) \bigg] = \prod_{i=1}^{n}b_i$$ over any choice of real constants $a_1, a_2, .....a_n$ and $b_1, b_2, ....b_n$.
Let us consider first an example with only $n=3$ random variables. So, we will have $X_{1},X_{2},X_{3}$ are a multiplicative system.
And we will prove that
$$\mathbb{E}[\prod_{i=1}^{3}(a_{i}X_{i}+b_{i})] =\prod_{i=1}^{3}b_{i}$$
We have the product $(a_{1}X_{1}+b_{1})*(a_{2}X_{2}+b_{2})*(a_{3}X_{3}+b_{3})$, where if we expand it we will have that it is equal to
$$a_{1}a_{2}a_{3}X_{1}X_{2}X_{3}+b_{2}a_{1}a_{3}X_{1}X_{3}+b_{3}a_{1}a_{2}X_{1}X_{2}+b_{2}b_{3}a_{1}X_{1}+\\b_{1}a_{2}a_{3}X_{2}X_{3}+b_{1}b_{3}a_{2}X_{2}+b_{1}b_{2}a_{3}X_{3}+b_{1}b_{2}b_{3} \ \ \ (**)$$
We results in a summation of $8$ products. However, only one of these products doesn't include a subset of the $X_{1},X_{2},X_{3}$ and this is the $b_{1},b_{2},b_{3}$.
Hence, from the definition of the Multiplicative System we will have that $\mathbb{E}[X_{1}X_{2}X_{3}]=\mathbb{E}[X_{1}X_{3}]=\mathbb{E}[X_{1}X_{2}]=\mathbb{E}[X_{1}]=\mathbb{E}[X_{2}X_{3}]=\mathbb{E}[X_{2}]=\mathbb{E}[X_{3}]=0$
So, by taking the expected value over $(**)$ we have that the only term that has non-zero expected value is $\mathbb{E}[b_{1}b_{2}b_{3}]=\prod_{i=1}^{3}b_{i}$.
Now, let consider the case were we have $n=4$ random variables, again if we expand the product $\prod_{i=1}^{4}(a_{i}X_{i}+b_{i})$ it will result into a summation of products, where only one product will not include a subset of $X_{1},X_{2},X_{3},X_{4}$ and that will be the $b_{1}b_{2}b_{3}b_{4}$. In that kind of analogy in the case where we have $X_{1},X_{2},...,X_{n}$ random variables, in the expantion of the product $\prod_{i=1}^{n}(a_{i}X_{i}+b_{i})$ there will be only one term that it is free of any subset of $X_{1},X_{2},...,X_{n}$ and that would be the $\prod_{i=1}^{n}b_{i}$.
