<s>[INST] How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$? Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
 [/INST] "Combining" the answers from Qiaochu Yuan and Isaac one can also directly evaluate 
$\begin{bmatrix}x \\ y \end{bmatrix} = e^{\mathbf{J}t} \begin{bmatrix}1 \\ 0 \end{bmatrix}$
with $\mathbf{J}=\begin{bmatrix}0&1 \\ -1&0 \end{bmatrix}$ (note that $\mathbf{J}^2 = - \begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}$) as follows:
$$
e^{\mathbf{J}t} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \left( \mathbf{J}^0 + \frac{t}{1!} \mathbf{J}^1 + \frac{t^2}{2!} \mathbf{J}^2 + \frac{t^3}{3!} \mathbf{J}^3 + \frac{t^4}{4!} \mathbf{J}^4 + \cdots \right) \begin{bmatrix}1 \\ 0 \end{bmatrix} \\ = \left( \begin{bmatrix}1&0 \\ 0&1 \end{bmatrix} + t \begin{bmatrix}0&1 \\ -1&0 \end{bmatrix} - \frac{t^2}{2!}\begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}-\frac{t^3}{3!}\begin{bmatrix}0&1 \\ -1&0 \end{bmatrix}+ \frac{t^4}{4!}\begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}+\cdots\right)\begin{bmatrix}1 \\ 0 \end{bmatrix} \\ = \begin{bmatrix} 1 + 0 - \frac{t^2}{2!} - 0 + \frac{t^4}{4!} + \cdots \\ 0 - t - 0 + \frac{t^3}{3!} + 0 + \cdots   \end{bmatrix} \\ = \begin{bmatrix} \cos(t) \\ - \sin(t) \end{bmatrix}
$$
The result is a parametrization of a helix curve. That means Eulers formula simply shows how one can parametrize a helix using the exponential function. 
The result can also be written like
$$
e^{\mathbf{J}t} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \left( \cos(t) + \mathbf{J} \sin(t) \right)\begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
making the connection to the traditional form of Eulers formula manifest.
It should be noted that with $\mathbf{K}=\begin{bmatrix}0&1 \\ 1&0 \end{bmatrix}$ the following relation can be proved in the same way:
$$
e^{\mathbf{K}t} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \left( \cosh(t) + \mathbf{K} \sinh(t) \right)\begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
</s>