First of all, given the matrix $A$, due to primary decomposition theorem we can restrict our view to a single generalized eigenspace $V_{\lambda} := Ker(A-\lambda Id)^{n_{\lambda}}$,where $n_{\lambda}$ is the algebraic multiplicity of $\lambda$ in the characteristical polynomial $P_{A}(t)$.
In this case the job is much easier, since we can work just with one eigenvalue and repeat this process with all of them.
Secondly we are going to do a further semplification; we additionally restrict our view to a single Jordan block of size $n$ with eigenvalue $\lambda \ne 0$.
Now we are able to determine the Jordan form of $J^{k}$, $\hspace{0.1cm} k \in \mathbb{Z}$. In other words we can easily answer the question as long as we have a single Jordan block of a given size.
We just need the following :
$Proposition :$ Let $A$ be the upper triangular matrix
$$ A = \begin{pmatrix}\mu & a_{1,2} & a_{1,3} & \cdots & a_{1,n} \\ & \mu & a_{2,3} & \cdots & a_{2,n} \\ & & \ddots & \ddots & \vdots \\ & & & \mu & a_{n-1,n} \\ & & & & \mu\end{pmatrix}$$
With $a_{1,2}\cdot a_{2,3}\cdots \cdots a_{n-1,n} \ne 0$, then the jordan form of $A$ is
$$J_{\mu,n} = \begin{pmatrix} \mu & 1 & & & \\ & \mu & \ddots \\
& & \ddots \\ & & & & \mu \end{pmatrix}$$
Proof of Proposition :
The matrix $A$ has just one eigenvalue $\mu$ of algebraic multiplicity $n$.
Moreover, $A-\mu Id$ is nilpotent and is index of nilpotencty is $n$. Since the index of nilpotency of $A-\mu Id$ is the biggest size of a block in Jordan form associated to the eigenvalue $\mu$ in the Jordan form of $A$,
It follows that the Jordan of $A$ has just one block of order $n$, hence it's $J_{\mu,n} \hspace{0.2cm}\Box$.
Now we can answer our original problem. For the assumpution made
(Note that this is not exactly the Jordan of $A$, so the notation $J$ is improper, this would represent the Jordan form of a block of size $n$ related to the eigenvalue $\lambda$,one of whom we restricted our view)
now we have:
$$J = \begin{pmatrix} \lambda & 1 & & & \\ & \lambda & \ddots \\
& & \ddots \\ & & & & \lambda \end{pmatrix}$$
By induction on $k$ we can see that
$$ J^{k} = \begin{pmatrix}\lambda^{k} & k\lambda^{k-1} & \\ & \lambda^{k} & \ddots & \\ & & \ddots & k\lambda^{k-1} \\ & & & \lambda^{k}\end{pmatrix}$$
So, $J^{k}$ is an upper triangular matrix with eigenvalue $\lambda^{k}$ of algebraic multiplicity $n$, with non-zero elements on the above parallel diagonal (here we are using $\lambda \ne 0$).
Thanks to the proposition, the Jordan form of $J^{k}$ has only one block of size $n$, which means it is $J_{\lambda^{k},n}$.
Edit : Using induction we can at most extend this argument for every $k \in \mathbb{N}$.
Since we want to work even with $k=-1$ we can use the following result to determine negative powers :
Why does the $n$-th power of a Jordan matrix involve the binomial coefficient? taking $f(x) = x^{k}$.
Now it works with the same argument for every value of $k \in \mathbb{Z}$.
Note : If $\lambda = 0$, it no longer holds that the Jordan form of $J^{k}$ has just one block.
Let's take $n=3,k=2$ as example with
$$J = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \hspace{0.5cm} J^{2} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
The jordan form of $J^{2}$ is :
$$J(J^{2}) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$
Observe that if $k=n, J^{n} = 0$.
