1. Assume for contradiction: Suppose there exists a positive integer $n$ such that for $k = 1, 2, \ldots, 9$, the leftmost digit of $(n+k)!$ equals $k$.

2. Define the reduced copy: Let the reduced copy of a positive integer $n$ be $a$, where $n = a \cdot 10^b$ for a nonnegative integer $b$ and a terminating decimal $a$ such that $1 \le a < 10$.

3. Leading digit condition: If the leading digit of $n$ is $k$, then the reduced copy of $n$ is between $k$ and $k+1$.

4. Analyze the reduced copy of factorials:
- The reduced copy of $(n+1)!$ is between $1$ and $2$.
- The reduced copy of $(n+2)!$ is between $2$ and $3$.

5. Generalize the bounds:
- For $(n+2)!$, the reduced copy must be between $\frac{2}{2} = 1$ and $\frac{3}{1} = 3$.
- For $(n+3)!$, the reduced copy must be between $\frac{3}{3} = 1$ and $\frac{4}{2} = 2$.
- For $(n+9)!$, the reduced copy must be between $\frac{9}{9} = 1$ and $\frac{10}{8} = 1.25$.

6. Establish the range: Note that the lower bound is always $1$ and the upper bound starts at $3$ and slowly reduces to $1.25$.

7. Claim: The reduced copy of all numbers from $n+2$ to $n+9$ must be between $1$ and $1.25$.

8. Proof by contradiction:
- Suppose not. Since the numbers from $n+2$ to $n+9$ are consecutive, there will be a number from $n+2$ to $n+9$ with a reduced copy between $9$ and $10$.
- This is impossible because that does not fit into any of the earlier ranges (which had a lower bound of $1$ and an upper bound of at most $3$).

9. Contradiction:
- Note that the reduced copy of $(n+1)!$ is between $1$ and $2$.
- Since the reduced copies of $n+2, n+3, n+4$ are all between $1$ and $1.25$, the reduced copy of $(n+4)!$ is less than $2(1.25)^3 = 3.90625 < 4$.
- This is a contradiction as it must be between $4$ and $5$.

Therefore, there is no positive integer $n$ such that for $k = 1, 2, \ldots, 9$, the leftmost digit of $(n+k)!$ equals $k$.