$p_n = 1- \left( 1-\frac{1}{365}\right)\left( 1-\frac{2}{365}\right)\cdots \left( 1-\frac{n-1}{365}\right)$ Then $p_n>\frac{1}{2}$ for $n>?$
This occured in a probability problem. The result was simply stated that this is true for $n>25$. I think I can see that this expression is monotonically increasing. But I am not sure that differentiating an expression like this which is not in closed form is correct. For example, I did
$\frac{dp_n}{dn} = + \frac{1}{365}\left( 1-\frac{1}{365}\right)\left( 1-\frac{2}{365}\right)\cdots \left( 1-\frac{n-2}{365}\right)$
But I still have $n-2$ so is this derivative correct? Otherwise I do not see how this is even monotonic. What would be the correct way to use calculus for this so that I can find $n$ after which the expression is more than half. If there are any algebraic techniques I would be glad to know them.
This came from: $p_n = 1-\frac{^{365}P_n}{365^n}$ where $^nP_x$ is the number of x-permutations of n objects .