The correct answer is $\boxed{\text{A. At least 15}}$.
The minimal polynomial of a square matrix $A$ is the monic polynomial of least degree $n$ such that $A^n – r_1 A^{n-1} + \cdots + (-1)^n r_n I = 0$ for some scalars $r_1, r_2, \dots, r_n$. The eigenvalues of $A$ are the roots of the minimal polynomial.
Since $A$ is a square matrix of order 100, it has 100 eigenvalues. If $A$ has exactly 15 distinct eigenvalues, then the minimal polynomial must have at least 15 roots, which means that its degree must be at least 15.
It is possible for the minimal polynomial to have a higher degree than 15, but this is not always the case. For example, if $A$ is a diagonal matrix with 15 distinct diagonal entries, then the minimal polynomial is simply the product of the linear factors $(x – r_1), (x – r_2), \dots, (x – r_{15})$, which has degree 15.
Therefore, the degree of the minimal polynomial of a square matrix of order 100 that has exactly 15 distinct eigenvalues is at least 15.