Something is wrong with $P_n$...
You seem to be trying to intuit the answer from the start, and an error has occurred somewhere along the way. It is much safer in probability to make smaller steps and arrive at the solution, rather than coming up with an intuitive answer.
Rather than directly thinking about the probability that no person shares the same birthday, it is safer to first formulate the underlying space, then find the probability we care about by describing it as a set in the space.
The following is going to seem over-formal, and for this level of problem it probably is. But all of the steps will be clear and it will show a safe way of thinking about this sort of question.
We could say that each person's birthday is a random variable $X$ which is an identity function on an event space. We will say $X:\Omega\rightarrow \Omega$ and: $$\Omega = \{\text{birthdays someone could have}\}= \{\text{Jan. 1},\dots, \text{Dec. 31}\} =\{\omega_1,\dots, \omega_{365}\}.$$
We also assume that if $\omega \in \Omega$ then $P(\omega)=\frac{1}{365}.$
Using this notation, $P(X=\omega)=\frac{1}{365}$ if $\omega \in \{\omega_1,\dots, \omega_{365}\}$.
Furthermore, if $A\subseteq \Omega,$ then $P(X\in A)=\frac{|A|}{\Omega}.$
Now say we have some sequence of people $X_1,\dots,X_n$ with independent birthdays. We can form the space of their birthdays as:
$$\Omega^n=\{\text{possible birthdays }n\text{ people can have}\}= \underset{n}{\underbrace{\Omega \times \dots \times \Omega}}.$$
If $\omega_1,\dots,\omega_n \in \Omega$, by independence:
$$P((X_1,\dots,X_n)=(\omega_1,\dots,\omega_n)) = \underset{n}{\underbrace{\frac{1}{365}\cdot \dots \cdot \frac{1}{365}}}=\frac{1}{365^n}.$$
So then if $A\subseteq \Omega^n$:
$$P((X_1,\dots,X_n)\in A)=\frac{|A|}{365^n}$$
Now we can formulate our original question in terms of these random variables: $$P(\{\text{No one shares the same birthday}\})=P(X_1\neq X_2\neq X_3\neq \dots \neq X_n)$$
From this we can write the probability we care about in terms of a set $S$ in $\Omega^n$:
$$S=\{(\omega_1,\dots,\omega_n):\text{every }\omega_i\text{ is distinct}\}$$
And now to find the probability we care about we just have to find the size of that set. If you think about it, you will see that the size of the set is 'the number of ways we can choose $n$ different things from a set of 365 distinct objects when order matters.'
The first one we can choose any of the $365$, the next one we can choose anything except the first one we chose, i.e. $364$ and so on, so we have:
$$|S|=\underset{n}{\underbrace{365\cdot \dots \cdot (365-n+1)}}=\frac{365!}{(365-n)!}$$
Finally, we have $P(S)=|S|/365^n=\frac{\frac{365!}{(365-n)!}}{365^n}.$
