Is there any simple way to solve $366^n(366-n)!\geq 2\times 366!$?

Question

I am teaching probability for junior high school students. On the last page there is a "challenge yourself" section asking

What is the least number of students in a classroom for the probability that at least two of them have their birthday falling on the same day of the year to be greater than $1/2$?

It leads to an inequality $366^n(366-n)!\geq 2\times 366!$. I have solved it with program as follows.

The answer is 23 students.

Question

I wonder whether there is another simpler way to solve it for junior high school students with pencil and paper only.

Answer 1

The arithmetic-geometric inequality tells us $$\frac{N!}{(N-n)!}\le\left(N-\frac{n-1}{2}\right)^n $$ and is quite sharp when $n\ll N$. Thus a good approximation for $n$ might be the solution of $$N^n=2\cdot \left(N-\frac{n-1}{2}\right)^n,$$ or: $$\left(1-\frac{n-1}{2N}\right)^n \approx \frac 12$$ With $c:=n^2/N$ and for $n\gg 1$, $$\left(1-\frac{n-1}{2N}\right)^n\approx\left(1-\frac c{2n}\right)^n\approx e^{-\frac c2}.$$ This suggests $c\approx 2\ln 2$ and so $n\approx \sqrt{2N\ln 2}$. With $N=366$, this crude approximation gives us $n\approx 22.53$.