The probability person $B$ shares person $A$'s birthday is $1/N$, where $N$ is the number of equally possible birthdays,
so the probability $B$ does not share person $A$'s birthday is $1-1/N$,
so the probability $n-1$ other people do not share $A$'s birthday is $(1-1/N)^{n-1}$,
so the expected number of people who do not have others sharing their birthday is $n(1-1/N)^{n-1}$,
so the expected number of people who share birthdays with somebody is $n\left(1-(1-1/N)^{n-1}\right)$.
The basic idea was right, and a small modification is enough.
Line up the people in some arbitrary order. There are, under the usual simplifying assumption that the year has $365$ days, $365^{23}$ possible birthday sequences. Under the usual assumptions of independence, and that all birthdays are equally likely, all these sequences are equally likely. The assumption "equally likely" is not correct, though it is more correct for people than for eagles.
Now we count how many ways we can have precisely $2$ people have the same birthday, with everybody else having a different birthday, meaning different from each other and also different from the birthday of our birthday couple.
The couple can be chosen in $\binom{23}{2}$ ways. For each of these ways, the couple's birthday can be chosen in $365$ ways. And the birthdays of the others can be chosen in what is sometimes called $P(364,21)$ ways. (I have always avoided giving it a name.) So the number of birthday assignments that satisfy our condition is
$$\binom{23}{2}(365)P(364,21), \quad\text{that is,}\quad \binom{23}{2}(365)(364)(363)\cdots (344).$$
For the probability, divide by $(365)^{23}$.
Remark: Or else argue this way, which may be closer in spirit to your thinking. Pick two people, $i$ and $j$. What is the probability these two have the same birthday, and all other birthdays are different from this one, and different from each other? Whatever $i$'s birthday is, the probability that $j$'s matches it is $\frac{1}{365}$. By the usual birthday argument, the probability that the birthdays of the other people are different, and different from the birthdays couple's, is
$$\frac{364}{365}\frac{363}{365}\frac{363}{365}\cdots\frac{344}{365}.$$
Multiply the above expression by $\frac{1}{365}$.
Finally, sum the result over the $\binom{23}{2}$ ways to choose $i$ and $j$, that is, multiply by $\binom{23}{2}$.
Best Answer
Here is a possible approach which seems to gets most of the way to that expression:
Let $F_\lambda(x)=P(X\le x)$ be the cumulative distribution function for a Poisson distribution with mean $\lambda$. We will have $F_\lambda(x-1) \lt 1 - P(X=x)=1-e^{-\lambda} \frac{\lambda^x}{x!}$ but this will be close for $x \gg \lambda$.
Then for an individual day, the probability of fewer than $k$ birthdays falling on that day has a binomial expression but can be approximated with a Poisson expression using $\lambda=\frac{n}{c}$. So we might say it could be about $1-e^{-n/c} \dfrac{(n/c)^k}{k!}$.
Making the slightly wrong assumption that the number of birthdays falling on each day is almost independent of the number on other days, the probability they are all below $k$ might be about $\left(1-e^{-n/c} \dfrac{(n/c)^k}{k!}\right)^c$ which, using $\left(1-\frac{b}{m}\right)^m \approx \exp(-b)$, might be about $\exp\left(-e^{-n/c}\dfrac{n^k}{c^{k-1} k!} \right)$. Subtract this from $1$ to get an approximation to the probability that out of $n$ people at least $k$ birthdays occur on a single day in a year with $c$ days.
This not quite the same expression as you quoted, as it has $-1$ where your quote has $\left(\dfrac n{c(k+1)} -1\right)^{-1}$, but it has all the other elements.