Deriving a Quaternion Extension of Euler's Formula According to Wikipedia:

A rotation through an angle of $\theta$ around the axis defined by a unit vector
  $\vec{u} = (u_x, u_y, u_z) = u_x\mathbf{i} + u_y\mathbf{j} + u_z\mathbf{k}$
  can be represented by a quaternion. This can be done using an extension of Euler's formula:
$$ \mathbf{q} = e^{\frac{\theta}{2}{(u_x\mathbf{i} + u_y\mathbf{j} + u_z\mathbf{k})}} = \cos \frac{\theta}{2} + (u_x\mathbf{i} + u_y\mathbf{j} + u_z\mathbf{k}) \sin \frac{\theta}{2}$$

Yet this identity is never proven. How does one do so?
 A: The same way de Moivre's is proven for $i$. The square roots of $-1$ in the quaternions $\mathbb{H}$ are precisely the unit imaginary quaternions $u$. (Exercise: prove this.) Therefore,
$$\begin{array}{ll} \exp(\theta u) & \displaystyle =1+\theta u+\frac{\theta^2}{2!}u^2+\frac{\theta^3}{3!}u^3+\cdots \\ & \displaystyle =\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right)+\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots\right)u \\ & =\cos(\theta)+\sin(\theta)u. \end{array} $$
A: Since ${\vec{u}=u_x\mathbf{i}+u_y\mathbf{j}+u_z\mathbf{k}}$ is a unit vector of a 3D sphere by definition, $u^2=u_x^2\mathbf{i}^2+u_y^2\mathbf{j}^2+u_z^2\mathbf{k}^2 + (u_xu_y\mathbf{i}\mathbf{j} + u_yu_x\mathbf{j}\mathbf{i}) + (u_xu_y\mathbf{i}\mathbf{j} + u_yu_x\mathbf{j}\mathbf{i}) + (u_xu_y\mathbf{i}\mathbf{j} + u_yu_x\mathbf{j}\mathbf{i})$.
Having $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2 = -1$ and $\mathbf{i}\mathbf{j} = -\mathbf{j}\mathbf{i}$, $\mathbf{j}\mathbf{k}=-\mathbf{k}\mathbf{j}$, $\mathbf{k}\mathbf{i}=-\mathbf{i}\mathbf{k}$, $\vec{u^2}=-(u_x^2+u_y^2+u_z^2)=-1$. With this, Euler's formula holds as $i$ is replaced by $u$ and $\theta$ is replaced by $\frac{\theta}{2}$:
$$e^{\frac{\theta}{2}u} = \cos{\frac{\theta}{2}} + u\sin{\frac{\theta}{2}} $$
