For example, $1 k$ is a natural number, and $\frac{1001 \cdot 1002 \cdot \cdots \cdot 2005 \cdot 2006}{11^{k}}$ is an integer, what is the maximum value of $k$?
Solve $\frac{1001 \cdot 1002 \cdot \cdots \cdot 2005 \cdot 2006}{11^{k}}=\frac{2006!}{11^{k} \cdot 1000!}$.
The highest power of the prime factor 11 in 2006! is
$$
\left[\frac{2006}{11}\right]+\left[\frac{2006}{11^{2}}\right]+\left[\frac{2006}{11^{3}}\right]=182+16+1=199 \text {. }
$$
The highest power of the prime factor 11 in 1000! is
$$
\left[\frac{1000}{11}\right]+\left[\frac{1000}{11^{2}}\right]=90+8=98 \text {. }
$$

Therefore, the maximum value of $k$ is $199-98=101$.