Calculating the matrix exponential of rotation matrix

Split the sum into odd and even terms

\begin{eqnarray} e^A &=& \sum_{k=0}^{+\infty}\frac{A^{2k}}{(2k)!} + \sum_{k=0}^{+\infty}\frac{A^{2k+1}}{(2k+1)!} \\ &=& \sum_{k=0}\frac{(-1)^{k}}{(2k)!}I + \sum_{k=0}^{+\infty}\frac{(-1)^{k}}{(2k+1)!}A \\ &=& \cos(1) I + \sin(1)A \\ \end{eqnarray}

You have\begin{align}e^A&=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&-1\\1&0\end{pmatrix}+\frac12\begin{pmatrix}-1&0\\0&-1\end{pmatrix}+\frac1{3!}\begin{pmatrix}0&1\\-1&0\end{pmatrix}+\frac1{4!}\begin{pmatrix}1&0\\0&1\end{pmatrix}+\cdots\\&=\begin{pmatrix}1-\frac1{2!}+\frac1{4!}-\cdots&-1+\frac1{3!}-\frac1{5!}+\cdots\\1-\frac1{3!}+\frac1{5!}-\cdots&1-\frac1{2!}+\frac1{4!}-\cdots\end{pmatrix}\\&=\begin{pmatrix}\cos(1)&-\sin(1)\\\sin(1)&\cos(1)\end{pmatrix}.\end{align}

A sketch:

First consider this matrix as a matrix in $\mathcal M_2(\mathbf C)$, and diagonalise it. Its characteristic polynomial is $\lambda^2+1$, so the eigenvalues are $i,-i$. One finds corresponding eigenvectors: $e_1=(1,1)$ and $e_2=(1,i)$. In this basis, the diagonal matrix $D$ of eigenvalues has exponential: $$\exp D=\exp\begin{pmatrix}\mathrm e^i &0\\0&\mathrm e^{-i}\end{pmatrix}.$$ Now, the change of basis matrix from the canonical basis to basis $(e_1,e_2)$ and its inverse are $$P=\begin{pmatrix}i&1\\1&i\end{pmatrix},\quad P^{-1}=\tfrac12\begin{pmatrix}-i&1\\1&-i\end{pmatrix}.$$ Also, we know that $D=P^{-1}AP,\quad \text{so }\enspace A=PDP^{-1}$and $$\exp A=P\exp(D)P^{-1}=\text{ (some computation) }= \begin{pmatrix}\cos 1&-\sin 1\\\sin 1&\cos 1\end{pmatrix}$$ i.e. the matrix of rotation by 1 rad.