For me, the most convincing argument is of differential nature.
Let us begin, for a 2D rotation, by the addition property:
$$R(\phi)R(d\phi)=R(\phi+d\phi)\tag{1}$$
which can be transformed into:
$$\dfrac{R(\phi+d\phi)-R(\phi)}{d\phi}=R(\phi)\dfrac{R(d\phi)-Id}{d\phi}\tag{2}$$
Accept for the moment that (2) expresses the fact that:
$$R'(\phi)=R(\phi).J \ \ \text{where} \ \ J:=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\tag{3}$$
(where the "prime" symbol means "derivative wrt $\phi$)
In (3), we recognize the differential equation for the classical (1D...) exponential. From there, it is rather normal to still denote the solution by an exponential. But there is more to it. Indeed, if we consider :
$$S(\phi):= \begin{pmatrix}\cos \phi&-\sin \phi\\ \sin \phi&\cos \phi\end{pmatrix}\tag{4}$$
and replace $R(\phi)$ by $S(\phi)$ in (3), we get an identity : $S(\phi)$ is a solution of differential equation (3) (and all other solutions being multiple of this one, it is the only solution such that $S(0)=Id$).
It remains, coming back to (2), to understand what the hell is this $J$ coming from...
Plainly, comparing the form of $J$ with (4), we see that it expresses a $+\pi/2$ rotation (which is, btw, the role devoted to multiplication by number $i$, hehe). Now the final step, why is
$$\dfrac{R(d\phi)-Id}{d\phi} \ \text{identifiable with} \ J \ ?$$
Just because:
$$\dfrac{R(d\phi)-Id}{d\phi}=\dfrac{1}{d\phi}\left(\begin{pmatrix}\cos d\phi&-\sin d\phi\\ \sin d\phi&\cos d\phi\end{pmatrix}-\begin{pmatrix}1&0\\0&1\end{pmatrix}\right)$$
has a first order Taylor expansion (recall we need first order only for the derivative, for example no $(d\phi)^2$ terms):
$$\dfrac{R(d\phi)-Id}{d\phi}=\dfrac{1}{d\phi}\begin{pmatrix}1+... -1 &d\phi+...\\ -d\phi+...&1...-1\end{pmatrix}$$
where we recognize indeed $J$ (alias, under its mask, complex number $i$)!
