Powers of matrices: if $A^3 = A$, how can we show that $A^{27}= A$? Suppose that $A^3 = A$. How can we show that $A^{27}= A$?
Any guidance would be much appreciated. 
Is it sufficient or correct to write that: $$A^9=(A^3)^3=(A)^3= A \implies  A^{27}= (A^9)^3 = (A)^3=A ?$$
 A: Yep, it's sufficient and correct to write what you've written!
If you're in doubt, the reason why this works is that the matrix product $A\times B$ is an associative binary operation. This means that you can define $A^n$ for all integers $n\geq 1$ by induction, and you can prove that the usual identities hold: $A^{n+m}=A^nA^m$ and $A^{nm}=(A^n)^m$. Since the identity matrix $I$ is the identity element for matrix multiplication, you can even define $A^0=I$ and verify that the identities still hold. In fact, if $A$ has an inverse $A^{-1}$, then you can define $A^n$ for all integers $n$, and the identities still hold!
Edit: Let me expand on this, because there was briefly a misleading answer here. (It's since been deleted.) What is the relationship between the numbers $27$ and $3$ that allows us to conclude $A^3=A\implies A^{27}=A$? One easy relationship, which you've observed, is that $27$ is the third power of $3$, so we can apply the former equation three times to get the latter equation.
But there is another relationship, which is even more generally useful: $27-1=26$ is a multiple of $3-1=2$. That allows us to argue that $A^{27}=A^{25}=A^{23}=\cdots=A^3=A$. This is not a coincidence! Whenever an integer $b$ is a power of an integer $a$, so $b=a^n$, we have that $a-1$ divides $b-1$. In fact:
$$(b-1)=(a^n-1)=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1).$$
So you see, the power relationship follows from the divisibility relationship.
Finally, to address the misleading answer directly: Yes, it so happens that $27$ is divisible by $3$. However, that's a red herring. It's not generally sufficient for $b$ to be a multiple of $a$. Consider the case when $A=-1$ as a $1\times1$ matrix. Then $A^3=A$, but $A^6\neq A$.
A: Does this help?
$$A^{27}=((A^3)^3)^3$$
