 Find the largest $n$ such that the last nonzero digit of $n!$ is $1$.

1. We start by noting that the factorial of a number \( n \), denoted \( n! \), is the product of all positive integers up to \( n \). Specifically, \( n! = n \times (n-1) \times (n-2) \times \cdots \times 1 \).

2. The last nonzero digit of \( n! \) is influenced by the factors of 2 and 5 in the factorial. This is because every pair of factors 2 and 5 contributes a trailing zero in the decimal representation of \( n! \).

3. For \( n = 0 \), we have \( 0! = 1 \). The last nonzero digit is 1.

4. For \( n = 1 \), we have \( 1! = 1 \). The last nonzero digit is 1.

5. For \( n \geq 2 \), \( n! \) will always include at least one factor of 2. Since \( n! \) is the product of all integers from 1 to \( n \), and 2 is included in this product, \( n! \) will be even for \( n \geq 2 \).

6. To determine the last nonzero digit of \( n! \) for \( n \geq 2 \), we need to consider the factors of 2 and 5. The presence of these factors will introduce trailing zeros, and we need to look at the remaining product after removing these trailing zeros.

7. For \( n = 2 \), we have \( 2! = 2 \). The last nonzero digit is 2.

8. For \( n = 3 \), we have \( 3! = 6 \). The last nonzero digit is 6.

9. For \( n = 4 \), we have \( 4! = 24 \). The last nonzero digit is 4.

10. For \( n = 5 \), we have \( 5! = 120 \). The last nonzero digit is 2 (after removing the trailing zero).

11. Continuing this process, we observe that for \( n \geq 2 \), the last nonzero digit of \( n! \) is not 1.

12. Therefore, the largest \( n \) such that the last nonzero digit of \( n! \) is 1 is \( n = 1 \).

The final answer is \( \boxed{1} \).