A wizard thinks of a number from $1$ to $n$. You can ask the wizard any number of yes/no questions about the number. The wizard must answer all those questions, but not necessarily in the respective order. What is the least number of questions that must be asked in order to know what the number is for sure. (In terms of $n$.)

Fresh translation.
1. **Claim**: The least number of questions required to determine the wizard's number for sure is \( n-1 \).

2. **Achievability**: 
   - We can ask the following sequence of questions:
     1. Is the number \( 1 \)?
     2. Is the number \( 1 \) or \( 2 \)?
     3. Is the number \( 1 \), \( 2 \), or \( 3 \)?
     ...
     \( n-1 \). Is the number \( 1 \), \( 2 \), ..., or \( n-1 \)?
   - If we get \( x \) yes answers, where \( 0 < x \leq n-1 \), then we know the wizard's number is \( x \), because exactly \( x \) questions will give a response of yes.
   - If the wizard answers no to each of the questions, then his number is \( n \).

3. **Proof by Contradiction**:
   - Assume for contradiction that it is possible to use fewer than \( n-1 \) questions to determine the wizard's number.
   - This means that if the wizard gives \( x \) yes responses, where \( 0 \leq x \leq n-2 \), then we can determine his number.
   - Therefore, each \( x \) corresponds to at most one number between \( 1 \) and \( n \).
   - However, there are \( n \) possible numbers and at most \( n-1 \) possible values of \( x \).
   - This leads to a contradiction because we cannot map \( n \) numbers to \( n-1 \) values uniquely.
   - Hence, the least number of questions is indeed \( n-1 \).

\(\blacksquare\)

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