There are $ n \plus{} 1$ cells in a row labeled from $ 0$ to $ n$ and $ n \plus{} 1$ cards labeled from $ 0$ to $ n$. The cards are arbitrarily placed in the cells, one per cell. The objective is to get card $ i$ into cell $ i$ for each $ i$. The allowed move is to find the smallest $ h$ such that cell $ h$ has a card with a label $ k > h$, pick up that card, slide the cards in cells $ h \plus{} 1$, $ h \plus{} 2$, ... , $ k$ one cell to the left and to place card $ k$ in cell $ k$. Show that at most $ 2^n \minus{} 1$ moves are required to get every card into the correct cell and that there is a unique starting position which requires $ 2^n \minus{} 1$ moves. [For example, if $ n \equal{} 2$ and the initial position is 210, then we get 102, then 012, a total of 2 moves.]
1. **Base Case:**
   Consider the base case where \( n = 1 \). There are 2 cells labeled 0 and 1, and 2 cards labeled 0 and 1. The objective is to get card 0 into cell 0 and card 1 into cell 1. If the initial position is 10, we need one move to get to 01. This satisfies the condition \( 2^1 - 1 = 1 \).

2. **Inductive Hypothesis:**
   Assume that for \( n = k \), the maximum number of moves required to sort the cards is \( 2^k - 1 \). We need to show that for \( n = k + 1 \), the maximum number of moves required is \( 2^{k+1} - 1 \).

3. **Inductive Step:**
   Suppose \( n = k + 1 \). We need to show that the number of steps it takes for card \( k+1 \) to move to position \( k+1 \) is at most \( 2^{k+1} - 1 \).

   - If card \( k+1 \) is already in position \( k+1 \), no moves are needed for this card, and the problem reduces to sorting the remaining \( k \) cards, which by the inductive hypothesis can be done in \( 2^k - 1 \) moves.
   
   - If card \( k+1 \) is in position \( h \) where \( h < k+1 \), we need to move it to position \( k+1 \). The allowed move is to find the smallest \( h \) such that cell \( h \) has a card with a label \( k > h \), pick up that card, slide the cards in cells \( h+1, h+2, \ldots, k \) one cell to the left, and place card \( k \) in cell \( k \).

4. **Detailed Calculation:**
   - If in fewer than \( 2^{k-1} \) steps, card \( k+1 \) moves forward, then we are done by induction.
   - If this is not the case, then in \( 2^{k-1} \) steps, all changes have occurred only in positions \( 0 \) through \( k-1 \). Hence, it follows that \( \pi(0), \pi(1), \ldots, \pi(k-1) \in \{0, 1, \ldots, k-1\} \) and hence \( \pi|_{\{0,1,\ldots,k-1\}} \in \mathcal{S}_k \). By induction, we know that these can be sorted in at most \( 2^k - 1 \) steps, and so card \( k+1 \) can be transferred to position \( k+1 \) in at most \( 2^k \) steps.

5. **Conclusion:**
   Therefore, the total number of moves required to sort \( k+1 \) cards is at most \( 2^k + 2^k - 1 = 2^{k+1} - 1 \).

6. **Uniqueness of Maximum Moves:**
   Clearly, if the number of steps is to be maximum, then again we can argue inductively that the corresponding permutation has to be \( 12\cdots(n-1)n1 \).

\(\blacksquare\)

The final answer is \( \boxed{ 2^n - 1 } \).