As 1mdlrjcmed said in a deleted answer, this is more the coupon collector's problem than the birthday problem. The expected number until you first see all days is $365H_{365} \approx 2364.646$ using harmonic numbers.
As Sassatelli Giulio said in a comment, you can in theory calculate your probability of a free day using Stirling numbers of the second kind, and you are looking for $1 - \frac{365!\, S_2(N,365)}{365^N}$.
I would suggest instead using recursion, where $p(N, m)$ is the probability of $m$ days taken with $N$ people. You want $1-p(N,365)$ and you can use $$p(N,m) = \frac{m}{365}p(N-1,m)+\frac{366-m}{365}p(N-1,m-1)$$ starting with $p(0,0)=1$ and $p(N,0)=p(0,m)=0$ when $N$ or $m$ are non-zero.
Clearly you will get $1-p(N,365)=1$ when $N<365$. You will get
- about $1-1.5\times 10^{-157}$ when $N=365$,
- about $0.99$ when $N=1606$
- about $0.9$ when $N=1853$
- about $0.5$ when $N=2287$
- about $0.1$ when $N=2971$
- about $0.01$ when $N=3827$
which if graphed looks like this