Is there any easy method of computing
$$A^{n}x$$ when $A$ is not diagonalizeable?
That is, we cannot find $P,D$ such that $A=PDP^{-1}$.
If there is no easy way of computing, can we say anything about the resulting quantity? For instance, convergence to the eigenvector with the largest eigenvalue?
From the wikipedia page on spectral radius, we know that
if $A=VJV^{-1}$ is the Jordan canonical form, where
$$J=\begin{bmatrix} J_{m_1}(\lambda_1) & 0 & 0 & \cdots & 0 \\ 0 & J_{m_2}(\lambda_2) & 0 & \cdots & 0 \\ \vdots & \cdots & \ddots & \cdots & \vdots \\ 0 & \cdots & 0 & J_{m_{s-1}}(\lambda_{s-1}) & 0 \\ 0 & \cdots & \cdots & 0 & J_{m_s}(\lambda_s) \end{bmatrix}$$
and $$J_{m_i}(\lambda_i)=\begin{bmatrix} \lambda_i & 1 & 0 & \cdots & 0 \\ 0 & \lambda_i & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_i & 1 \\ 0 & 0 & \cdots & 0 & \lambda_i \end{bmatrix}\in \mathbf{C}^{m_i \times m_i}, 1\leq i\leq s.$$
Then we have $$A^k=VJ^kV^{-1}$$
and
$$J^k=\begin{bmatrix} J_{m_1}^k(\lambda_1) & 0 & 0 & \cdots & 0 \\ 0 & J_{m_2}^k(\lambda_2) & 0 & \cdots & 0 \\ \vdots & \cdots & \ddots & \cdots & \vdots \\ 0 & \cdots & 0 & J_{m_{s-1}}^k(\lambda_{s-1}) & 0 \\ 0 & \cdots & \cdots & 0 & J_{m_s}^k(\lambda_s) \end{bmatrix}$$
and
$$J_{m_i}^k(\lambda_i)=\begin{bmatrix} \lambda_i^k & {k \choose 1}\lambda_i^{k-1} & {k \choose 2}\lambda_i^{k-2} & \cdots & {k \choose m_i-1}\lambda_i^{k-m_i+1} \\ 0 & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} & \cdots & {k \choose m_i-2}\lambda_i^{k-m_i+2} \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} \\ 0 & 0 & \cdots & 0 & \lambda_i^k \end{bmatrix}$$
If the spectral radius is less than $1$, then the quantity converges to zero.
From your description about convergence to the dominant eigenvector, don't forget to normalize your matrix product in every iteration of power iteration. Even in the case of unique dominant eigenvalue, the convergence is not promised.