I have seen the solution of the paradox on Wikipedia, but before that I tried it out myself. But I got my answer wrong with my approach.
To calculate the probability of no two people in a group of 23 having the same birthday, I divided {the number of ways 23 different days in 365 days can be taken} by {the number of ways 22,21,20,...,2,1 different days can be taken multiplying with the number of ways 23 people can have 22,21,20,...,2,1 birthdays respectively}
I think that would make it a bit harder to understand what I wanted to convey. And, I find using MathJax a bit tough as I'm new to this. So the link provides the image of what I tried on desmos:My expression of what I tried
You have written $$1-\dfrac{{365\choose 23}}{\sum\limits_{i=1}^{23} i^{23-i}{365 \choose i}}$$ presumably as the probability that there are at least two people sharing a birthday, with $i$ iterating over the possible numbers of distinct birthdays.
Both the numerator and the denominator raise issues. The numerator is a combination count but if you are counting equally likely events then you want permutations: if there are two people, it is twice as likely that one has a birthday on January 1 and another on January 2 compared with both having birthdays on January 1. So you should have $\frac{365!}{(365-23)!}$ as the numerator.
The other problem is the denominator. It gives the correct number when $i=1$ of $365$ but not for other $i$. For example when $i=2$ it gives $2^{22} {365 \choose 2}$ when the number of ways of allocating $23$ people between two days with both used is $(2^{23}-2){365 \choose 2}$, almost twice as high. But you do not need a sum over $i$ anyway: the number of equally likely ways of allocating $23$ birthdays (possibly duplicated) over $365$ days is $365^{23}$.
Putting these together means the probability of at least two people out of $23$ sharing a birthday is $$1-\dfrac{\frac{365!}{(365-23)!}}{365^{23}} \approx 0.5073$$ i.e. just over a half.