Answer
See below
Work Step by Step
$A=\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0& 0&0 \end{bmatrix} \\ A^2=\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0& 0&0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0& 0&0 \end{bmatrix} =\begin{bmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0& 0&0 \end{bmatrix}$ $A^3=A^2.A=\begin{bmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0& 0&0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0& 0&0 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0& 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0& 0&0 \end{bmatrix}\\ A^4=\begin{bmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 & 0& 0&0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 &0 \\ 0 & 0 & 0 & 1 \\ 0 & 0& 0&0 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 & 0& 0&0 \end{bmatrix}$
Hence, an $n \times n$ matrix $A$ is nilpotent.
$e^{At}=I_4+At+\frac{1}{2!}(At)^2+\frac{1}{3!}(At)^3+\frac{1}{4!}(At)^4...+\frac{1}{k!}(At)^k\\ =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\0 & 0& 0 & 1 \end{bmatrix} +\begin{bmatrix} 0 & t & 0 & 0 \\ 0 & 0 & t & 0\\ 0 & 0 & 0& t\\0 & 0& 0 & 0\end{bmatrix}+\frac{1}{2}\begin{bmatrix} 0 & 0 & t^2 & 0\\ 0 & 0 & 0 & t^2\\ 0& 0 & 0 & 0\\ 0& 0 & 0 & 0 \end{bmatrix} +\frac{1}{6}\begin{bmatrix} 0 & 0 & 0& t^3\\ 0 & 0 & 0 & 0\\ 0& 0 & 0 & 0\\ 0& 0 & 0 & 0 \end{bmatrix}\\ =\begin{bmatrix} 1 & t &\frac{1}{2}t^2 & \frac{1}{6}t^3\\ 0 & 1 & t & \frac{1}{2}t^2 \\ 0 & 0 & 1 & t\\ 0 & 0 & 0 &1 \end{bmatrix}$
