Do matrices have a "to the power of" operator? Well I was sure that saying "$A^3$"  (where $A$ is an $n\times n$ matrix) is nonsense. Sure one could do $(A\cdot A) A$  But that contains different operators etc. So what did my prof mean by the following statement:

show that $A^{25}\mathbf{x} = \mathbf{0}$ has only the trivial solution? (We're also given the determinant of A).

I know the proof will probably end with stating: "This means that $A^{25}$ is invertible, so $A^{25}\mathbf{x} = \mathbf{0}$ has only the trivial solution. And well I could state that $\det(A^{25}) = 5^{25} \neq 0$.

But then again: I really wonder what the "to the power of" operator means? Or did my prof make a mistake here?
If you want a formal definition of matrix exponentiation for non-negative integer values, just define $A^n = A^{n-1}\cdot A$ and $A^0 = I$. Since matrix multiplication is associative, we won't have any ambiguity there.

Edit: As Tobias Kildetoft points out below, it might be wiser to define the base case as  $A^1=A$ instead of $A^0=I$, so as to not have to worry about how $\det(A^n)=\det(A)^n$ for $\det(A)=0$ would imply $0^0=1$. Which isn't false, depending on how we want to define it, but is something we might not want to worry about for the purposes of defining matrix powers.
But matrix multiplication is not commutative, $A^{2}\cdot A^{3} \neq A^{3}\cdot A^{2}$ except in some particular cases.
$$
    A = \pmatrix{0 &amp; 3 \\ 1&amp;2}; A^2 = \pmatrix{3 &amp; 6 \\ 2 &amp; 7}; A^3 = \pmatrix{6&amp;  21 \\ 9 &amp; 20}
$$
$$A^2·A^3 = \pmatrix{72 &amp; 183\\ 75 &amp; 182}; A^3·A^2 = \pmatrix{60 &amp; 183 \\ 61 &amp; 182}
$$
So $A^5$ has more than 1 solution, at least in this case.