The approaches in the OP already seem quite straightforward to me, and as is clear from the answer by Arthur B, doing the calculation exactly is not too complicated.
Anyway, here is my suggestion (after a bit of scribbling) for how one might have seen the solution at a glance.
All we have to do is compare the winning chances of person $k$ and person $k+1$ (provided we do it for general $k$). Edit: this is because the winning chances are first increasing and then decreasing, which of course isn't clear a priori, so perhaps the previous sentence should have started "As it turns out...".
We imagine two persons A and B who are to occupy places $k$ and $k+1$. We only have to compare the scenarios where the one who goes first wins, to the scenarios where the one who goes second wins (otherwise the order of A and B doesn't matter).
Those scenarios first of all require that persons $1,\dots, k-1$ have different birthdays.
Under that assumption, the cases where the person to go first wins are when both A and B have a birthday which is already represented among the first $k-1$ people. Counting combinatorially rather than probabilistically, there are $(k-1)^2$ ways of choosing the birthdays of A and B with that constraint.
The cases where the person to go second wins are when A and B have the same birthday, and that birthday is not shared with any of the first $k-1$ people. There are $n-(k-1)$ such ways of choosing the birthdays of A and B.
Therefore we end up comparing $(k-1)^2$ to $n-(k-1)$, or equivalently comparing $k(k-1)$ to $n$.
After having written it down, I get the feeling that this argument is more convoluted than the straightforward calculation by Arthur B, but here it is, for what it's worth.