recall the distributive law for sets: $\displaystyle A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
Now, let $\displaystyle P(n)$: $\displaystyle (A_1 \cap A_2 \cap \cdots \cap A_n) \cup B = (A_1 \cup B) \cap (A_2 \cup B) \cap \cdots \cap (A_n \cup B)$ for all $\displaystyle n \in \mathbb{N}, ~n \ge 1$.
$\displaystyle P(1)$ is trivially true.
Assume $\displaystyle P(n)$ is true. We show $\displaystyle P(n + 1)$
Define $\displaystyle C = A_1 \cap A_2 \cap \cdots \cap A_n$
then, $\displaystyle (A_1 \cap A_2 \cap \cdots \cap A_n \cap A_{n + 1}) \cup B = (C \cap A_{n + 1}) \cup B = (C \cup B) \cap (A_{n + 1} \cup B)$ By the distributive law.
since $\displaystyle P(n)$ is true. $\displaystyle (C \cup B) = (A_1 \cup B) \cap (A_2 \cup B) \cap \cdots (A_n \cup B)$
so that we have $\displaystyle (A_1 \cap A_2 \cap \cdots \cap A_n \cap A_{n + 1}) \cup B = (A_1 \cup B) \cap (A_2 \cup B) \cap \cdots \cap (A_{n + 1} \cup B)$
so that $\displaystyle P(n + 1)$ is true.
This completes the inductive proof