Proof by induction with an nxn-matrix The matrix A $ \in \mathbb{R}^{n\times n}$ is of the form
\begin{equation*}
A = \begin{pmatrix}
0 & 1 & 0 & \cdots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
 &  & \ddots & \ddots & 0  \\  
\vdots &  &  & \ddots & 1  \\
0 & \cdots &  & \cdots & 0  \\ 
\end{pmatrix}
\end{equation*}
Now I want to compute $e^{tA}$ and $e^{tA} = \sum_{k=0}^{\infty} \frac{1}{k!}\cdot (tA)^{k}$.
I observed that $A^{2}$ is equal to the matrix A only with de "diagonal" of ones moved 1 up.
\begin{equation*}
A^{2} = \begin{pmatrix}
0 & 0 & 1 & \cdots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
 &  & \ddots & \ddots & 1  \\  
\vdots &  &  & \ddots & 0  \\
0 & \cdots &  & \cdots & 0  \\ 
\end{pmatrix}
\end{equation*}
And $A^{3}$ is equal to the matrix A with the "diagonal" of ones moved 2 up.
So to compute $e^{tA}$ I want to proof that $A^{n}$ is equal to the matrix A with the "diagonal" of ones moved up n-1 times, which results in $A^{n} = 0$. I have tried to proof it with induction.
So claim: If A $ \in \mathbb{R}^{n\times n}$ is of the above form, then $A^{n} = 0$.
For n = 2, the 2 x 2 matrix is equal to:
\begin{equation*}
A = 
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}
\end{equation*}
And $A^{2} = 0$.
So the claim holds for n = 2.
But I don't know how to do the induction step in this case.
Other ways to compute $e^{tA}$ with this matrix A are also welcome.

Prove the following by induction on $n$:
Lemma: Let $A$ be an $m \times m$  matrix of any size such that
$$A_{ij}= \left\{ 
\begin{array}{lc}
1 &\mbox{ if } j-i=1 \\
0 & \mbox{ otherwise}
\end{array}
\right.
$$
Then, for all $n$ we have
$$(A)^n_{ij}= \left\{ 
\begin{array}{lc}
1 &\mbox{ if } j-i=n \leq m \\
0 & \mbox{ otherwise}
\end{array}
\right.
$$
The inductive step is immediate
$$
(A)^{n+1}_{ij}=(A)^n_{i1}A_{1j}+(A)^n_{i2}A_{2j}+...+(A)^n_{im}(A)^n_{mj}
$$
Just note here that
$$
(A)^n_{ik}A_{kj}=0
$$
unless $k-i=n$ and $j-k=1$.
Alternate proof
Just observe that $A$ is the matrix corresponding to the Linear Transformation $T: \mathbb R^m \to \mathbb R^m$.
$$
T(e_k)=\left\{ 
\begin{array}{lc}
e_{k+1} &\mbox{ if } k<m \\
0 & \mbox{ if } k=m
\end{array}
\right.
$$
Prove by induction on $n$ that
$$
T^n(e_k)=\left\{ 
\begin{array}{lc}
e_{k+n} &\mbox{ if } k+n \leq m \\
0 & \mbox{ if } k+n>m
\end{array}
\right.
$$
