We say "$s$ grows to $r$" if there exists some integer $n>0$ such that $s^n = r.$ Call a real number $r$ "sparcs" if there are only finitely many real numbers $s$ that grow to $r.$ Find all real numbers that are sparse.
1. We start by analyzing the condition \( s^n = r \). This implies that \( s = \sqrt[n]{r} \) for some integer \( n > 0 \).

2. We need to determine for which real numbers \( r \) there are only finitely many real numbers \( s \) such that \( s^n = r \).

3. Consider the case when \( r = 1 \):
   - For \( n = 1 \), \( s = \sqrt[1]{1} = 1 \).
   - For \( n = 2 \), \( s = \sqrt[2]{1} = \pm 1 \).
   - For any \( n \), \( s = \sqrt[n]{1} = 1 \) or \( s = -1 \) if \( n \) is even.
   - Therefore, there are only two real numbers \( s \) (namely \( 1 \) and \( -1 \)) that satisfy \( s^n = 1 \).

4. Consider the case when \( r = -1 \):
   - For \( n = 1 \), \( s = \sqrt[1]{-1} = -1 \).
   - For \( n = 3 \), \( s = \sqrt[3]{-1} = -1 \).
   - For any odd \( n \), \( s = \sqrt[n]{-1} = -1 \).
   - Therefore, there is only one real number \( s \) (namely \( -1 \)) that satisfies \( s^n = -1 \).

5. Consider the case when \( r = 0 \):
   - For any \( n \), \( s = \sqrt[n]{0} = 0 \).
   - Therefore, there is only one real number \( s \) (namely \( 0 \)) that satisfies \( s^n = 0 \).

6. Now, consider the case when \( r \) is any real number other than \( -1, 0, \) and \( 1 \):
   - For each \( n \), there exists a real number \( s = \sqrt[n]{r} \).
   - For example, if \( r > 0 \), \( s = \sqrt[n]{r} \) is a real number for any \( n \).
   - If \( r < 0 \), \( s = \sqrt[n]{r} \) is a real number for any odd \( n \).
   - Therefore, there are infinitely many real numbers \( s \) that satisfy \( s^n = r \).

7. From the above analysis, we conclude that the only real numbers \( r \) for which there are finitely many real numbers \( s \) that grow to \( r \) are \( -1, 0, \) and \( 1 \).

The final answer is \(\boxed{-1, 0, 1}\).