You are my complement

$P(A^{C} \cap B^{C}) = P((A \cup B)^{C}) = 1 - P(A \cup B) =$ $= 1 - (P(A) + P(B) - P(A \cap B)) =$ Now, A and B are independent events so $= 1 - (P(A) + P(B) - P(A) \cdot P(B)) = 1 - P(A) - P(B) + P(A) \cdot P(B) =$ $= (1 - P(A)) \cdot (1 - P(B)) = P(A^{C}) \cdot P(B^{C})$

A proof without a mathematical treatment would be to take the example of tossing of a coin.

Events "head occurs" & "tail occurs" are independent of each other, so is their complements i.e. "tail occurs" & "head occurs".

Another good example in this context would be rolling of a standard die.