I learn from
Wikipedia that a quaternion can be written generally as
q = {x_0} + i{x_1} + j{x_2} + k{x_3},
where {x_0}, {x_1}, {x_2}, {x_3} are real numbers. And {i^2} = {j^2} = {k^2} = - 1
Its conjugation is written,
{q^*} = {x_0} - i{x_1} - j{x_2} - k{x_3}.
Its norm can be written,
\left\| q \right\| = \sqrt {q{q^*}} = \sqrt {{x_0}^2 + {x_1}^2 + {x_2}^2 + {x_3}^2}.
But
q can also be written as a scalar plus a vector as,
q = {x_0} + \vec v,
where,
\vec v = i{x_1} + j{x_2} + k{x_3}.
Then the exponential of a quaternion,
q, can be written as,
{e^q} = \sum\limits_{n = 0}^\infty {\frac{{{q^n}}}{{n!}} = {e^{{x_0}}}(\cos \left\| {\vec v} \right\|} + \frac{{\vec v}}{{\left\| {\vec v} \right\|}}\sin \left\| {\vec v} \right\|).
So since the {\vec v} carries the
i, j, k components, it does indeed seem that the exponential of a quaternion is a quaternion.
I assume the same construction leads to the fact that the exponential of a octonion is an octonion.