Let $n$ be an integer. We consider $s (n)$, the sum of the $2001$ powers of $n$ with the exponents $0$ to $2000$. So $s (n) = \sum_{k=0}^{2000}n ^k$ . What is the unit digit of $s (n)$ in the decimal system?
1. We start with the given sum \( s(n) = \sum_{k=0}^{2000} n^k \). This is a geometric series with the first term \( a = 1 \) and common ratio \( r = n \).

2. The sum of a geometric series \( \sum_{k=0}^{m} r^k \) is given by the formula:
   \[
   s(n) = \sum_{k=0}^{2000} n^k = \frac{n^{2001} - 1}{n - 1} \quad \text{for} \quad n \neq 1
   \]

3. To find the unit digit of \( s(n) \), we need to consider the unit digit of \( \frac{n^{2001} - 1}{n - 1} \).

4. First, observe that for any integer \( n \), the unit digit of \( n^5 \) is the same as the unit digit of \( n \). This is because the powers of any digit repeat every 4 cycles in the units place (by Euler's theorem or direct observation).

5. Therefore, the unit digit of \( n^{2001} \) is the same as the unit digit of \( n \) because \( 2001 \mod 4 = 1 \).

6. Let \( d \) be the unit digit of \( n \). Then the unit digit of \( n^{2001} \) is \( d \).

7. Now, consider the expression \( \frac{n^{2001} - 1}{n - 1} \). We need to find the unit digit of this expression.

8. Since the unit digit of \( n^{2001} \) is \( d \), we have:
   \[
   n^{2001} \equiv d \pmod{10}
   \]
   Thus,
   \[
   n^{2001} - 1 \equiv d - 1 \pmod{10}
   \]

9. We need to divide \( d - 1 \) by \( n - 1 \) modulo 10. Since we are only interested in the unit digit, we can simplify the problem by considering the possible values of \( d \) (0 through 9) and their corresponding \( n - 1 \).

10. For each \( d \), we check the unit digit of \( \frac{d - 1}{n - 1} \). However, since \( n \) and \( d \) are arbitrary, we can generalize the result by noting that the sum of the geometric series modulo 10 will always yield a unit digit of 1, as the series sums to a number ending in 1 when \( n \neq 1 \).

11. Therefore, the unit digit of \( s(n) \) is 1.

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