The geometric interpretation of quaternion multiplication is fundamentally 4-dimensional (unlike quaternion conjugation, which can be considered as an action on $\Bbb{R}^3$).
Let's start with an easy case. Say $q=a+bi$ with $b \neq 0$, $a^2+b^2=1$. That is, $q$ is a non-real unit quaternion in the subalgebra of $\Bbb{H}$ generated by $i$. What affect does multiplying by $q$ have on an arbitrary quaternion $r$?
First of all, if $r$ also lies in the subalgebra generated by $i$, then we can consider multiplication of $q$ and $r$ to be ordinary complex multiplication; that is, multiplication by $q$ rotates $r$ by $\theta$ in the $\{1, i\}$ plane.
Secondly, if $r$ lies in the orthogonal complement of that subalgebra, $r=cj+dk$, we can write $r=(c+di)j=j(c-di)$. The first of these representations can be used to left-multiply by $q$ via ordinary complex multiplication; the second one can be used to right-multiply. In either case, multiplying by $q$ rotates $r$ by $\theta$ in the $\{j, k\}$-plane; however, the sign difference means that the two multiplications rotate in opposite directions from each other.
We can then find the effect of multiplying $q$ by an arbitrary quaternion by projecting that quaternion into these two planes. That is, an arbitrary quaternion will have its $\{1, i\}$-projection and $\{j, k\}$-projection both rotated by $\theta$ when it is multiplied by $q$; the direction of the $\{j, k\}$-rotation, but not of the $\{1, i\}$-rotation, will be affected by whether we're left-multiplying or right-multiplying by $q$.
The general case works similarly. For any unreal unit quaternion $q$ that makes an angle $\theta$ with the real axis, multiplication by $q$ rotates by $\theta$ in the $\{1, q\}$-plane, and also rotates by $\theta$ in its orthogonal complement. The direction of the first rotation is fixed, but the direction of the second rotation depends on whether we're multiplying by $q$ on the left or on the right. You can see this just by noticing that any unreal quaternion generates a 2-dimensional subalgebra isomorphic to $\Bbb{C}$, making the previous few paragraphs work in general after some relabeling.
This also gives you a way to see why quaternion conjugation works the way it does on $\Bbb{R}^3$. If $q$ makes an angle $\theta$ with the real axis, then the map $r \mapsto qrq^{-1}$:
- is the identity map in the $\{1, q\}$-plane, since that plane forms a commutative subalgebra of $\Bbb{H}$
- involves a rotation by $2\theta$ in its orthogonal complement. Both $q$ and $q^{-1}$ rotate quaternions in $\{1, q\}^{\perp}$ by $\theta$. If they were multiplied on the same side, those rotations would have to cancel out; since left-multiplication behaves oppositely to right-multiplication, this means they must reinforce each other. Since the orthogonal complement of $\{1, q\}$ is orthogonal to $1$, it is pure imaginary, so we've reproduced the fact that quaternion conjugation corresponds to a double-angle rotation in $\Bbb{R}^3$ (when identified with $\Im(\Bbb{H})$).
Note also that multiplying by a general quaternion involves scaling by the norm of that quaternion, but conjugation conveniently causes the norms of $q$ and $q^{-1}$ to cancel.
What would the formula be to take a quaternion to a scalar power n?
You'd need some elementary definitions for that.
Let $p=a+bi+cj+dk\in\mathbb{H}$ be a quaternion. Define, conjugacy (1), vector part (2), sign (3) and argument (4), as:
$$\overline{p}=a-bi-cj-dk\,\,\,\,\text{(1)}$$
$$\vec{u}=\frac{p-\overline{p}}{2}\,\,\,\,(2)$$
$$sgn(p)=
\begin{cases}
\frac{p}{|p|}, & \text{if $p\neq 0$ (3)} \\
0, & \text{if $p=0$}
\end{cases}$$
$$\arg(p)=
\begin{cases}
\arccos\left(\frac{a}{|p|}\right), & \text{if $p\neq 0$ (4)} \\
undefined, & \text{if $p=0$}
\end{cases}$$
Exponential and logarithmic functions can now be defined, because $\mathbb{H}$ has a division algebra, so
$$\exp(p)=\exp(a)\cdot(\cos(|\vec{u}|)+sgn(\vec{u})\sin(|\vec{u}|))\,\,\,\,\text{(5)}$$
and
$$\ln(p)=\ln(|p|)+sgn(\vec{u})\arg(p)\,\,\,\,\text{(6)}$$
Your scalar power will now be:
$$q^n=\exp(n\ln(q))\,\,\,\,(7)$$
(7) now allows you to calculate directly.
Note that in this case, $n$ is a scalar, so $n\ln(q)=\ln(q)n$, so scalar powers in this case are unique modulo the branch you are working in.
Best Answer
There is no point in trying to generalise the base $x$ of exponentiation $x^y$ to be a quaternion, since already for $x$ a complex number (other than a positive real) and non-integral rational $y$ there is no unique natural meaning to give to $x^y$. (For instance $z^{2/3}$ could be interpreted as asking for the square of a cube root of$~z$, or for the cube root of $z^2$, and in both cases there are not one but (the same) three candidates; an attempt to force a single outcome for instance by fixing a preferred cube root for every complex number would make the two interpretations differ for certain$~z$.) Anyway, if anything $x^y$ is going to be equivalent to $\exp(\ln(x)y)$ or $\exp(y\ln(x))$ (giving you some choice in case of non-commutatvity) for some meaning of $\ln x$. So the whole effect of using a strange $x$ is to multiply the exponent by a constant; one is better off just writing that multiplication explicitly and sticking to the exponential function $\exp$.
There is no problem at all to extend $\exp$ to a function $\Bbb H\to\Bbb H$, by the usual power series. In fact every non-real quaternion spans a real subalgebra isomorphic to$~\Bbb C$, which will be $\exp$-stable, and restricted to it $\exp$ will behave just as the complex exponential function. Of course one can only expect $\exp(x+y)=\exp(x)\exp(y)$ to hold if $\exp(x)$ and $\exp(y)$ commute, which essentially is the case when $x$ and $y$ lie in the same subalgebra isomorphic to$~\Bbb C$ (and hence commute).