How to compute the $n^{\textrm{th}}$ power of a matrix?

Imagine that $A$ is diagonalizable, i.e, you can write $A = P^{-1}DP$ for some $P$ and diagonal $D$. Then, $$ A^2 = \left( P^{-1}DP \right) \left( P^{-1}DP \right) = \left( P^{-1}D \left(P P^{-1}\right) DP \right) = \left( P^{-1}D^2 P \right) $$

Can you take this from here?

EDIT In response to the comments below, I am getting $$ P = \pmatrix{1 & -1\\-1 & 2}, P^{-1}= \pmatrix{2 & 1\\1 & 1}, D= \pmatrix{-1 & 0\\0 & 3} $$ so $$ A^5 = \pmatrix{2 & 1\\1 & 1} \pmatrix{-1 & 0\\0 & 243} \pmatrix{1 & -1\\-1 & 2} = \pmatrix{-245 & 488\\-244 & 487} $$

It may help to know that the diagonal entries of $D$ are the eigenvalues of $A$, and the rows of $P$ are the corresponding eigenvectors of $A$, so calculate the eigenvalues and eigenvectors, then $A^5=(P^{-1}D^5P)$

To find the diagonal form of the matrix you would calculated the eigenvalues and eigenvectors. $D$ is a diagonal matrix where the entries are the eigenvalues of $A$, $P$ is a matrix formed by the eigenvectors (each eigenvector is a column corresponding to the column its eigenvalue is in for $D$).