, 2002

Solution:

We assume that there exists a number $n$ with the required property. Then none of the factorials can be a power of ten, because from $3!$ onward all factorials are divisible by $3$ and the first few factorials obviously do not have the required property. Also, none of the numbers $n+2, \ldots, n+9$ can be a power of ten, because otherwise the leading digit of two consecutive factorials in the sequence under consideration would be the same. Thus there is a $j$ such that $10^{j} < n+2 < \ldots < n+9 < 10^{j+1}$ (1).

Because $(n+8)!$ begins with an $8$ and $(n+9)!$ begins with a $9$, there are natural numbers $a$ and $b$ with $9 \cdot 10^{a} < (n+9)! < 10^{a+1}$ and $8 \cdot 10^{b} < (n+8)! < 9 \cdot 10^{b}$, which leads to $10^{a-b} < n+9 < \frac{5}{4} \cdot 10^{a-b}$.

With (1) it follows that $j = a-b$ and $10^{j} < n+2 < \ldots < n+9 < \frac{5}{4} \cdot 10^{j}$ (2).

Since $(n+1)!$ begins with $1$, there is an $m$ such that $10^{m} < (n+1)! < 2 \cdot 10^{m}$, while from (2) it follows that: $10^{3j} < (n+2)(n+3)(n+4) < \left(\frac{5}{4}\right)^{3} \cdot 10^{3j}$.

Multiplying the last two inequalities gives $10^{3j+m} < (n+4)! < 2 \cdot \frac{125}{64} \cdot 10^{3j+m}$, which, because $\frac{250}{64} < 4$, can be weakened to $10^{3j+m} < (n+4)! < 4 \cdot 10^{3j+m}$. From this it follows that the number $(n+4)!$ would not begin with $4$, but with $1$, $2$ or $3$ - contradicting the assumption.