The direct approach, although not recommended, is do-able, via Inclusion-Exclusion. This approach provides a remedy to the issue of over-counting raised by the comment of Henry.
See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Let $~Q~$ denote the set of all possible distributions of the birthdays of the $~n~$ people, without any regard for whether any of the people share a birthday with you.
Let $~S~$ denote the subset of $~Q~$ where at least one person shares a birthday with you.
Then the desired computation of the probability is
$$\frac{|S|}{|Q|} ~: ~|Q| = (365)^n,$$
So, the problem reduces to computing $~|S|,~$ which represents the number of elements in the set $~S.$
Index the $~n~$ people $~P_1, P_2, \cdots, P_n.$
For $~k \in \{1,2,\cdots,n\},~$ let
$~S_k~$ denote the subset of $~S~$ such that person $~P_k~$ shares a birthday with you. For example, $~S_1~$ represents the set of all possible birthday distributions, where person $~P_1~$ shares a birthday with you, and any of the other people may or may not share a birthday with you.
Then, since $~( ~S_1 \cup S_2 \cup \cdots \cup S_n ~) = S,~$ the problem reduces to computing
$$| ~S_1 \cup S_2 \cup \cdots \cup S_n ~|.$$
Let $~T_1~$ denote $ ~|S_1| + |S_2| + \cdots + |S_n|.$
For $~r \in \{2,3,\cdots,n\},~$
let $~T_r~$ denote
$\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq n} | ~S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r} ~|.$
That is, $~T_r~$ represents the sum of $~\displaystyle \binom{n}{r}~$ terms.
Then, in accordance with Inclusion Exclusion theory,
$$| ~S_1 \cup S_2 \cup \cdots \cup S_n ~|
= \sum_{r=1}^n (-1)^{r+1} T_r.$$
So, the problem reduces to computing $~T_1, T_2, \cdots, T_n.~$
$\underline{\text{Computation of} ~T_1}$
$S_1~$ represents the set where $~P_1~$ shares your birthday, and all of the other people are unrestricted.
Therefore, $~|S_1| = (365)^{n-1}.~$
By Symmetry, for each $~k \in \{2,3,\cdots,n\},~$ you have that $~\displaystyle |S_k| = |S_1|.~$
Therefore,
$$T_1 = \binom{n}{1} (365)^{n-1}.$$
$\underline{\text{Computation of} ~T_2}$
$( ~S_1 \cap S_2 ~)~$ represents the set where $~P_1~$ and $~P_2~$ both share your birthday, and all of the other people are unrestricted.
Therefore, $~|S_1 \cap S_2| = (365)^{n-2}.~$
By Symmetry, for each
$~1 \leq i_1 < i_2 \leq n ~: ~i_1, i_2 \in \Bbb{Z},$
you have that
$~\displaystyle |S_{i_1} \cap S_{i_2}| = |S_1 \cap S_2|.~$
Therefore,
$$T_2 = \binom{n}{2} (365)^{n-2}.$$
$\underline{\text{Computation of} ~T_r ~: 3 \leq r \leq n}$
The analysis in this section will parallel the analysis of the previous section.
There will be $~\displaystyle \binom{n}{r}~$ terms,
and each term will equal $~(365)^{n-r}.$
Therefore,
$$T_r = \binom{n}{r} (365)^{n-r}.$$
$\underline{\text{Final Computation}}$
$$\forall r \in \{1,2,\cdots, n\}, ~T_r = \binom{n}{r} (365)^{n-r}.$$
$$ |S| = \sum_{r=1}^n (-1)^{r+1} ~T_r.$$
The probability is
$$\frac{|S|}{|Q|} ~: ~|Q| = (365)^n.$$
$\underline{\text{Addendum}}$
By binomial expansion, the numerator of the final answer represents
$$(365)^n - \left[ ~365 - 1 ~\right]^n.$$
This implies that the final computation is
$$\frac{(365)^n - (364)^n}{(365)^n}. \tag1 $$
The expression in (1) above makes perfect sense, since $~(364)^n~$ represents the number of ways that the $~n~$ people can each avoid sharing your birthday.
