$3, \cdots, 2^{n-1}$, then the last number taken out $L(n-1)$ satisfies $2 L(n-1)=L(n)$, and $L(1)=2, L(2)=4, \cdots$, hence $L(n)=2^{n}$.

Arrange $1,2,3, \cdots, 2^{n}$ in a circle, and sequentially take out 1,3, $5, \cdots$, taking one number every other, then the last number taken out is $2^{n}$.

Analysis: Suppose $1,2,3, \cdots, 2^{n}$ are finally taken out according to the rule as $L(n)$, then after taking out $1,3,5, \cdots,\left(2^{n}-1\right)$, the remaining numbers are $2,4,6, \cdots, 2^{n}$, a total of $2^{n-1}$ numbers. If they are renumbered 1,2,