This is merely a rephrasing of Henry's answer. It's probably utterly useless to give the same proof again in different words, so I'm making it community-wiki.
Recall the secretary problem: At each time $k$, you want to decide whether to pick candidate $k$ or not. You win if you the candidate you pick is the best of all $n$ candidates, and you want to maximize the probability of winning.
Now note that at any time $k$, if candidate $k$ is not the best among candidates $1, \dots, k$, then if you pick the candidate you lose immediately. (This candidate is not even best among those seen at that point; so cannot be best among all candidates.) So you will only pick a candidate if she's best among those seen until then. Thus, any optimal strategy will be of the form that does, at each time $k$:
If candidate $k$ is better than everyone else seen previously and [some additional conditions], pick candidate $k$ and stop.
The "[some additional conditions]" must depend only on information you have upto time $k$, and in particular only on $k$ and the relative order of candidates $1$ to $k$. But since you only care about whether you pick the best candidate or not, the only relevant factor of the relative order is who the best is among those seen at that point, and that you already know is candidate $k$. So the "[some additional conditions]" is some predicate $P(k)$ that depends only on $k$.
Now note that if you pick candidate $k$ (who is best among $1$ to $k$), then the probability of winning is $k/n$ (the probability that the best was among the first $k$). This is an increasing function of $k$, which means that if $k$ is good so is $k+1$ (i.e., if $P(k)$ is true then so should $P(k+1)$ be). So $P(k)$ is of the form $[k \ge t]$ i.e., the range of good $k$ is some interval $\{t, \dots, n\}$ for some $t$.
This is what you wanted proved. (And as Henry observed, for the special case $n=2$, both $t=1$ and $t=2$ work, but for larger $n$, you can prove that $t$ should be $n/e$.)