I am wondering if we can generalise Bayes' formula if we are conditioning upon multiple events? It seems to make logical sense from the definition of Bayes' formula in the single conditioning case, but I am unsure how to prove my "conjectured" generalisation.
In more formal terms, I am asking whether or not it is that true that the following identity holds: $$ P(A|B_1, \dots, B_n) = \frac{P(A \cap B_1 \cap \dots \cap B_n)}{P(B_1 \cap \dots \cap B_n)}$$
I would be grateful for any help here.
Yes, that is true.
If we let $C = B_1 \cap \dots \cap B_n$, then we have the following:
$$P(A|B_1, \dots, B_n) = P(A|B_1 \cap \dots \cap B_n) = P(A|C)$$
Now we can apply the version of Bayes' rule that you are already familiar with and then substitute in our definition of $C$ to find that:
$$ P(A|C) = \frac{P(A \cap C)}{P(C)} = \frac{P(A \cap B_1 \cap \dots \cap B_n)}{P(B_1 \cap \dots \cap B_n)}$$
This confirms what you speculate in your question.