Answer
$P_{n}=\frac{365-(n-1)}{365} P_{n-1}$
Work Step by Step
We look for a pattern. ( If we find one, using mathematical induction it should not present a problem to prove that the pattern is valid for all integers. We don't do this unless explicitly asked, though)
$P_{1}=\frac{365}{365}=1 $
$P_{2}=\frac{365}{365} \cdot \frac{364}{365}=\frac{364}{365} P_{1}=\frac{365-(2-1)}{365} P_{1} \\
P_{3}=\frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}=\frac{363}{365} P_{2}=\frac{365-(3-1)}{365} P_{2}$
From the suggested pattern, we conclude
$P_{n}=\frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdot \cdots \frac{365-(n-1)}{365}=\frac{365-(n-1)}{365} P_{n-1}$
