Find the number of zeros in which the decimal expansion of $ 2007!$ ends. Also find its last non-zero digit.
1. **Finding the number of trailing zeros in \(2007!\):**

   The number of trailing zeros in \(2007!\) is determined by the number of times \(10\) is a factor in \(2007!\). Since \(10 = 2 \times 5\), and there are always more factors of \(2\) than \(5\) in factorials, we only need to count the number of factors of \(5\).

   The number of factors of \(5\) in \(2007!\) can be found using the formula:
   \[
   \left\lfloor \frac{2007}{5} \right\rfloor + \left\lfloor \frac{2007}{25} \right\rfloor + \left\lfloor \frac{2007}{125} \right\rfloor + \left\lfloor \frac{2007}{625} \right\rfloor
   \]

   Calculating each term:
   \[
   \left\lfloor \frac{2007}{5} \right\rfloor = \left\lfloor 401.4 \right\rfloor = 401
   \]
   \[
   \left\lfloor \frac{2007}{25} \right\rfloor = \left\lfloor 80.28 \right\rfloor = 80
   \]
   \[
   \left\lfloor \frac{2007}{125} \right\rfloor = \left\lfloor 16.056 \right\rfloor = 16
   \]
   \[
   \left\lfloor \frac{2007}{625} \right\rfloor = \left\lfloor 3.2112 \right\rfloor = 3
   \]

   Summing these values:
   \[
   401 + 80 + 16 + 3 = 500
   \]

   Therefore, the number of trailing zeros in \(2007!\) is \(500\).

2. **Finding the last non-zero digit of \(2007!\):**

   To find the last non-zero digit of \(2007!\), we need to consider the factorial modulo \(10\) while ignoring the factors of \(10\) (i.e., factors of \(2\) and \(5\)).

   We use the fact that \(4! \equiv -1 \mod 5\) and apply it repeatedly. This is based on the property of factorials and modular arithmetic.

   First, we need to find \(2007!\) modulo \(5\):
   \[
   2007! \equiv 2 \mod 5
   \]

   This is because the product of numbers modulo \(5\) cycles every \(4\) numbers (since \(4! \equiv -1 \mod 5\)).

   Next, we need to consider the powers of \(2\) modulo \(10\). Since \(2007!\) has \(500\) trailing zeros, we can ignore these factors of \(10\) and focus on the remaining product modulo \(10\).

   By examining the pattern of the last non-zero digits of factorials, we find that the last non-zero digit of \(2007!\) is \(2\).

The final answer is \( \boxed{ 500 } \) trailing zeros and the last non-zero digit is \(2\).