Referring to the "Discussion" section of the following page, I wanted to know why P(B) wasn't $\frac{(n-1)^{k-1}}{n^{k-1}}$. Note $k$ is the number of people at a party and $n=365$.
Because there is 1 less person to account for, we need only consider $n^{k-1}$ permutations of possible birthdays for everyone.
Of those permutations, there is 1 less birthday date since we are accounted for. So the number of dates that may be birthday dates without matching my birthday is $(n-1)$. Since there are $(k-1)$ people at the party, the first person may have their birthday fall on any $(n-1)$ day, the second on any $(n-2)$, ..., the $k$'th on any $(n-k)$ day.
So the total number of possibilities for $(k-1)$ people to not have the same birthday is: $(n-1)(n-2)...(n-k)$ = $P^{n-1}_{k-1}$
Alright, I'll take a shot at this.
If there are $n$ "days" and $k-1$ people (besides yourself) in the room, then the probability that somebody has the same birthday as you is:
$$ 1 - \frac{(n-1)^{k-1}}{n^{k-1}} = 1 - \left(1-\frac 1n\right)^{k-1}$$
The proof is simple, we start by counting the opposite: the probability that nobody has the same birthday as you. For any person, this probability is $1 - \frac 1n$, and since there are $k-1$ people and the events are independent we have:
$$ P(\text{nobody has the same birthday}) = \left(1-\frac 1n\right)^{k-1} $$
We want the opposite of that event. This is why we take the complement by doing $1 - $ the above, and gives the stated answer.