Exponential of the matrix I want to calculate the matrix exponential $e^{tA}$ of the matrix with the first row being $(0,1)$ and the second $ (-1,0)$. It would be sufficient if you would me the most important steps.

Another approach to calculate the exponent $e^{tA}$ bases on the fact that this exponent solves the Cauchy problem
$$
\frac{d}{dt} Y(t) = AY(t), \quad t\in\Bbb R,\\Y(0)=I.
$$
Obviously, this problem splits to solving two Cauchy problems
$$
\frac{d}{dt} \begin{pmatrix}x(t)\\y(t)\end{pmatrix}  = A \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}y(t)\\-x(t)\end{pmatrix}, \quad t\in\Bbb R,\\x(0)=1,\quad y(0)=0\tag{1}.
$$
and
$$
\frac{d}{dt} \begin{pmatrix}x(t)\\y(t)\end{pmatrix}  = A \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}y(t)\\-x(t)\end{pmatrix}, \quad t\in\Bbb R,\\x(0)=0,\quad y(0)=1\tag{2}.
$$
Whatever method you use to solve problems (1) and (2) - passing to higher order equations, observing that $x^2(t)+y^2(t)=1$, etc -  you will arrive to solutions
$$\begin{pmatrix}x_1(t)\\y_1(t)\end{pmatrix}=\begin{pmatrix}\cos t\\-\sin t\end{pmatrix},\quad \begin{pmatrix}x_2(t)\\y_2(t)\end{pmatrix}=\begin{pmatrix}\sin t\\ \cos t\end{pmatrix}.$$
This allows to conclude that $$Y(t)=e^{tA} =  \begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}.$$
