One can define $e^x$, $\sin x$ and $\cos x$ in any Banach algebra by means of power series, e.g. $e^x = \sum x^n/n!$. Then, for any square root of $-1$ in the algebra (which we will denote by $i$, although there could be many, as there are in the quaternions), it is a formal consequence of the power series that $e^{ix}=\cos x + i \sin x$. Because these functions are defined via power series, this is all tautological, and there isn't really any deep significance.
There are things two that make Euler's formula interesting over the reals. First, $e^x$, $\cos x$ and $\sin x$ are usually defined by non-power series methods, and so Euler's formula expresses some sort of non-obvious relation. Second, these functions satisfy all sorts of identities which can be better understood in terms of the identity. However, when we are working over a noncommutative Banach algebra (like the quaternions or $n \times n$ matrices), we generally have that $e^{a+b}=e^a e^b$ only when $ab=ba$, and so the most useful algebraic property of $e^x$ is gone, which means that so are the interesting relations you might want.
Edit: I would like to explain the calculation done in
Robert's answer in terms of what I have written above.
Let $q=a+bi+cj+dk=a+r\frac{bi+cj+dk}{r}$ where $r^2=b^2+c^2+d^2$, and we have assumed $r>0$. Note that $\frac{bi+cj+dk}{r}$ is a square root of $-1$, which we will temporarily write as $\sqrt{-1}$. Since $a$ commutes with everything, we have
$$e^q=e^{a+r\sqrt{-1}}=e^a e^{r\sqrt{-1}} = e^a( \cos(r) + \sqrt{-1} \sin(r))= e^a(\cos(r) + \frac{\sin(r)}{r}(bi+cj+dk)).$$
So we see that everything follows from the standard formula $e^{ix}=\cos x + i \sin x.$
Why is it that when we convert radians to degrees we multiply radians $\times \frac{180}π$ , but when we convert slope per radians to slope per degrees we have to multiply the inverse conversion formula slope per radians $\times \frac{π}{180}$
If we want to know an hour in terms of minutes, we multiply 1 hour $\times \frac{60}{1}$, given the result in minutes. If we want to know how convert 180 minutes, we divide $180$ minutes by $60$, i.e., multiply $180 \times \frac 1{60}$.
You'll find this phenomenon in any conversion: To convert temperature in degrees Celsius to temp in Fahrenheit, we have $F = \frac 95 C + 32$. To convert to from F to C, we need to invert this: $C = \frac 59(F-32)$
$R \text{ radians}\;\times \dfrac{180^\circ}{\pi \;\text{radians}} = \dfrac{R\times 180^\circ}{\pi}$.
"Radians" cancels as the unit, leaving a numeric value expressed in degrees.
$D \text{ degrees}\; \times \dfrac{\pi \;\text{radians}}{180\; \text{degrees}} = \dfrac{D\pi\;}{180}\;\text{ radians}$.
"Degrees" cancels as the unit, leaving the value expressed in radians.
Moved from comments:
Note that slope per radian is a ratio: $\;\dfrac{\text{slope}}{\text{radians}}.\;$ So to obtain "slope per degree", you need to have $π$ radians in the numerator, to cancel the unit "radians" from the denominator, and $180$ degrees in the denominator, to end with slope/degrees.
Slope itself is not a "unit" per se, meaning it isn't a degree, or radian, a mm, or foot. It is unit-free: even if we assign distance units (meters, say) to displacement: e.g. $Δ\,y\text{m}=y_2\,\text{m}−y_1\,\text{m}$ and $Δ\,x\,\text{m}=x_2\text{m}−x_1\,\text{m},$ then we have $$\text{slope}\,= \dfrac{Δy\,\text{ m}}{Δx\,\text{m}}=\frac{Δy}{Δx},$$ you see that the units "m: meter" attached to "change in y" and "change in x" cancel in the ratio defining slope, leaving us with a unit-free scalar which slope really is.
Best Answer
Given that
$$180°=\pi$$
we could write
$$e^{180°i}=-1$$