Why can a matrix whose kth power is I be diagonalized? Say A is an n by n matrix over the complex numbers so that A raised to the kth power is the identity I. How do we show A can be diagonalized?
Also, if alpha is an element of a field of characteristic p, how do we show that the matrix A=[1, alpha; 0, 1] satisfies A raised to the pth power equals the identity I and cannot be diagonalized if alpha is nonzero.
Please be detailed. I really have no idea how to start on this one.

If you have the machinery of Jordan forms and/or minimal polynomials you can settle the questions with those. When working over $\Bbb{C}$ user152558's answer points at a useful direction, and Pedro's answer shows that this won't work over the reals.
Lacking such technology I proffer the following argument.
Remember that an $n\times n$ matrix (over a field $K$) can be diagonalized if and only if its eigenvectors span all of $K^n$ or, equivalently, if all the vectors in $K^n$ are linear combinations of eigenvectors.
Over $K=\Bbb{C}$ you can then do the following. Let $x\in K^n$ be arbitrary. Let $\omega=e^{2\pi i/k}$ be a primitive $k$th root of unity. The vectors
$$
z_j=\frac1k(x+\omega^{-j}Ax+\omega^{-2j}A^2x+\cdots+\omega^jA^{k-1}x),\quad j=0,1,\ldots,k-1,
$$
are then easily seen to be eigenvectors of $A$. Namely (You show this) 
$$
Az_j=\omega^jz_j.$$
Furthermore, because for all $j, 0<j<k$, we have $\sum_{t=0}^{k-1}\omega^{-jt}=0$ (You show this, too. Apply the formula for a geometric sum) we see that
$$
x=z_0+z_1+\cdots z_{k-1}.
$$
Therefore all the vectors of $K^n$ are linear combinations of eigenvectors, and we are done.
Note: this argument works whenever the field $K$ has a primitive $k$th root of unity and when it is possible to divide by $k$, i.e. the characteristic of $K$ is not a factor of $k$ (actually the latter property follows from the first, but let's skip that).
Your other question follows from the same characterization of diagonalizability. Using the characteristic polynomial you see that $1$ is the sole eigenvalue. But when $\alpha\neq0$, the corresponding eigenspace is 1-dimensional. Thus not all vectors of $K^2$ are linear combinations of eigenvectors and we are done.
