Let $\{\omega_1,\omega_2,\cdots,\omega_{100}\}$ be the roots of $\frac{x^{101}-1}{x-1}$ (in some order). Consider the set $$S=\{\omega_1^1,\omega_2^2,\omega_3^3,\cdots,\omega_{100}^{100}\}.$$ Let $M$ be the maximum possible number of unique values in $S,$ and let $N$ be the minimum possible number of unique values in $S.$ Find $M-N.$
1. **Identify the roots of the polynomial:**
   The roots of the polynomial \(\frac{x^{101} - 1}{x - 1}\) are the 101st roots of unity, excluding 1. These roots are given by:
   \[
   \omega_k = e^{2\pi i k / 101} \quad \text{for} \quad k = 1, 2, \ldots, 100.
   \]

2. **Define the set \(S\):**
   The set \(S\) is defined as:
   \[
   S = \{\omega_1^1, \omega_2^2, \omega_3^3, \ldots, \omega_{100}^{100}\}.
   \]

3. **Determine the maximum number of unique values \(M\):**
   To find the maximum number of unique values in \(S\), we need to consider the distinctness of the elements \(\omega_k^k\). Since \(\omega_k = e^{2\pi i k / 101}\), we have:
   \[
   \omega_k^k = \left(e^{2\pi i k / 101}\right)^k = e^{2\pi i k^2 / 101}.
   \]
   The values \(e^{2\pi i k^2 / 101}\) will be distinct if \(k^2 \mod 101\) are distinct. Since 101 is a prime number, the quadratic residues modulo 101 are distinct for \(k = 1, 2, \ldots, 100\). Therefore, \(M = 100\).

4. **Determine the minimum number of unique values \(N\):**
   To find the minimum number of unique values in \(S\), we need to consider the possibility of elements \(\omega_k^k\) being equal. This happens if:
   \[
   \omega_k^k = \omega_m^m \quad \text{for some} \quad k \neq m.
   \]
   This implies:
   \[
   e^{2\pi i k^2 / 101} = e^{2\pi i m^2 / 101} \quad \Rightarrow \quad k^2 \equiv m^2 \pmod{101}.
   \]
   Since 101 is a prime number, each non-zero residue modulo 101 has a unique inverse. Therefore, the minimum number of unique values occurs when all elements are equal, which is possible if all \(k^2 \mod 101\) are the same. This can be achieved by choosing \(k\) such that \(k^2 \equiv 1 \pmod{101}\). Hence, \(N = 1\).

5. **Calculate \(M - N\):**
   \[
   M - N = 100 - 1 = 99.
   \]

The final answer is \(\boxed{99}\).