Example 6 Arrange $1,2, \cdots, n$ in a clockwise circle. First, cross out 2, then continue to cross out every other number in a clockwise direction until only one number remains. Denote the last remaining number as $f(n)$. Find $f(n)$.
Solution: List the values of $f(1), f(2), \cdots$ as follows:
$\begin{array}{llllllllll}n: & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \cdots \\ f(n): 1 & 1 & 3 & 1 & 3 & 5 & 7 & 1 & 3 \cdots\end{array}$
From the above table, we can conjecture that the rule for the values of $f(n)$ is:
When $n=2^{k}$, $f(n)=f\left(2^{k}\right)=1$;
When $n=2^{k}+t$ (where $t=1,2, \cdots, 2^{k}-1$), $f(n)=2 t+1$.
This can be uniformly written as:
When $2^{k} \leqslant n<2^{k+1}\left(n=2^{k}+t, t<2^{k}\right)$,
$f(n)=2 t+1=2\left(n-2^{k}\right)+1$,

where $k=\left[\log _{2} n\right]$.
To prove the above conjecture, we first prove the following lemma:

Lemma 1: $f(2 n)=2 f(n)-1$. That is, when $n$ is even, $f(n)=2 f\left(\frac{n}{2}\right)-1$.

In fact, arrange the numbers $1,2, \cdots, 2 n$ in a circle in clockwise order. In the first round, some numbers are crossed out, leaving only $1,3,5, \cdots, 2 n-1$, a total of $n$ numbers. Re-number these $n$ numbers as $1,2, \cdots, n$, and continue to cross out some numbers according to the original rule until only one number remains. The new number of this remaining number is $f(n)$. Thus, the original number of this remaining number is $2 f(n)-1$ (substituting back the original variable). Therefore, $f(2 n)=2 f(n)-1$.

Lemma 2: $f(2 n+1)=2 f(n)+1$. That is, when $n$ is odd, $f(n)=2 f\left(\frac{n-1}{2}\right)+1$.
The proof is left to the reader.
Now, we prove the original problem by induction on $n$. When $n=1,2, \cdots, 9$, the conclusion holds by the table. Assume the conclusion holds for all natural numbers less than $n$, and consider the natural number $n$. Let $n=2^{k}+t$ (where $t<2^{k}$).
(1) If $t$ is even, by Lemma 1 and the induction hypothesis, we have
$$
\begin{aligned}
f(n) & =f\left(2^{k}+t\right)=2 f\left(\frac{2^{k}+t}{2}\right)-1 \\
& =2 f\left(2^{k-1}+\frac{t}{2}\right)-1 \\
& =2\left(2 \cdot \frac{t}{2}+1\right)-1=2 t+1 ;
\end{aligned}
$$
(2) If $t$ is odd, by Lemma 2 and the induction hypothesis, we have
$$
\begin{aligned}
f(n) & =2 f\left(\frac{2^{k}+t-1}{2}\right)+1 \\
& =2 f\left(2^{k-1}+\frac{t-1}{2}\right)+1 \\
& =2\left(2 \cdot \frac{t-1}{2}+1\right)+1=2 t+1 .
\end{aligned}
$$

Thus, the conjecture is proved.