If $A$ has complex entries and $A^m = I$, then $A$ is diagonalizable? I have consulted some answers on SE already, such as these two:
If $A^m=I$ then A is Diagonalizable
Show that if $A^{n}=I$ then $A$ is diagonalizable.
However, I am still confused. I understand that if we can show that $p_A (\lambda) = \lambda^n - 1$, then setting it to zero, we will have $(\lambda - 1)(\lambda^{n-1} + \cdots)=0$, so we have $n$ distinct roots. (or even simpler, root of unity, so it follows)
But how do we get $p_A (\lambda) = \lambda^n - 1$ in the first place? Cayley-Hamilton says that $A$ should satisfy its own characteristic polynomial. But in this case we don't have the characteristic polynomial to begin with?
 A: To show that $A$ is diagonalizable, it's sufficient to show that its minimal polynomial has distinct roots.  Since $A$ satisfies $A^m-I$, its minimal polynomial must divide $\lambda^m-1$, which has distinct roots.  So the minimal polynomial has distinct roots, meaning that $A$ is diagonalizable.
For a proof of the first sentence, see for example
Minimal polynomial and diagonalizable matrix
A: Here is a different argument that does not use characteristic or minimal polynomials. 
We have that $A$ is similar to its Jordan form, i.e., $A=SJS^{-1}$. Then
 $$ J^m=(S^{-1}JS)^m=A^m=I. $$
Now, to have $J^m=I$, we need that each Jordan block $J_k(\lambda)$ satisfies $J_k(\lambda)^m=I_k$. This can only happen when $k=1$ (see below), which forces $J$ to be diagonal, and so $A$ is diagonalizable. 

Here is the proof that $J_k(\lambda)^m\ne I$ if $k\geq2$. The case $\lambda=0$ is trivial because the diagonal entries will always be zero. When $\lambda\ne0$, we can prove by induction on $m$ that $[J_k(\lambda)^m]_{12}$ (the $1,2$ entry of $J_k(\lambda)^m$) is $m\lambda^{m-1}$. When $m=1$, we have $J_k(\lambda)_{12}=1=m\lambda^{m-1}$. Assume as inductive hypothesis that $[J_k(\lambda)^m]_{12}=m\lambda^{m-1}$ and $[J_k(\lambda)^m]_{rr}=\lambda^m$ for all $r$. Then 
\begin{align}
[J_k(\lambda)^{m+1}]_{12}&=[J_k(\lambda)\,J_k(\lambda)^{m}]_{12}
=\sum_{r=1}^n[J_k(\lambda)]_{1r}\,[J_k(\lambda)^m]_{r2}\\ \ \\
&=[J_k(\lambda)]_{11}\,[J_k(\lambda)^m]_{12}+[J_k(\lambda)]_{12}\,[J_k(\lambda)^m]_{22}\\ \ \\
&=\lambda\,m\lambda^{m-1}+1\,\lambda^m=(m+1)\lambda^m.
\end{align}
