The definition of quaternionic exponential is given by the absolutely convergent series
$$
e^z=\sum_{k=0}^\infty\dfrac{z^k}{k!}
$$
It is well known that, from this definition, if $x, y$ commute we have $e^xe^y=e^ye^x=e^{x+y}$.
Since real quaternions commute with all other quaternions, for $a \in \mathbb{R}$ we have $e^{a+z}=e^ae^z \; \forall z\in \mathbb{H}$ so, if $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$, we have $e^z=e^ae^\mathbf{v}$, where $\mathbf{v}$ is an imaginary (or vector) quaternion.
Now we have:

**claim**

If $ \mathbf{v} \in \mathbb{H}_P$ is an imaginary quaternion, putting $\theta=|\mathbf{v}|$ we have:
$$
e^\mathbf{v}= \cos\theta + \mathbf{v}\;\dfrac{\sin \theta}{\theta}
$$
**proof**

We note that:
$$
\mathbf{v}^2= (b \mathbf{i}+c \mathbf{j} +d \mathbf{k})(b \mathbf{i}+c \mathbf{j} +d \mathbf{k})=
-b^2-c^2-d^2=-|\mathbf{v}|^2
$$
so:
$$
\mathbf{v}^2= -\theta^2
\quad,\quad \mathbf{v}^3= -\theta^2\mathbf{v}
\quad,\quad \mathbf{v}^4= \theta^4
\quad,\quad \mathbf{v}^5= \theta^4 \mathbf{v}
\quad,\quad \mathbf{v}^6= -\theta^6
\quad,\quad \cdots
$$
and the series become.
$$
\begin{split}
e^\mathbf{v}&=\sum_{k=0}^\infty\dfrac{\mathbf{v}^k}{k!}=\\
%
&=1+\dfrac{\mathbf{v}}{1!}-\dfrac{\theta^2}{2!}-\dfrac{\theta^2\mathbf{v}}{3!}+\dfrac{\theta^4}{4!}+\dfrac{\theta^4\mathbf{v}}{5!}-\dfrac{\theta^6}{6!}+\cdots=\\
%
&=1+\dfrac{\theta\mathbf{v}}{1!\,\theta}-\dfrac{\theta^2}{2!}-\dfrac{\theta^3\mathbf{v}}{3!\,\theta}+\dfrac{\theta^4}{4!}+\dfrac{\theta^5\mathbf{v}}{5!\,\theta}-\dfrac{\theta^6}{6!}+\cdots=\\
%
&=\left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\dfrac{\theta^6}{6!}\cdots\right)+\dfrac{\mathbf{v}}{\theta}\left( \dfrac{\theta}{1!}-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}\cdots\right)=\\
%
&=\cos\theta +\dfrac{\mathbf{v}}{\theta}\sin\theta
\end{split}
$$

So the exponential of a quaternion is:
$$
e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right)
$$