Let me suggest the following story: if we consider $z = x+iy$ as a point in the complex plane then we have $z = (x,y)$ in the usual Cartesian coordinate notation. Since
$$ z=x+iy = x(1)+y(i) = x(1,0)+y(0,1)$$
we have $1=(1,0)$ and $i=(0,1)$ in this approach. Viewing $z=(x,y)$ as a vector based at the origin we have:
$$ \boxed{z = r\hat{r}} $$
where $\hat{r}$ is a unit-vector in the direction of increasing distance from the origin $r$. Again, we are viewing $z$ as a directed line-segment based at the origin.
Where is the polar form?
Let $\theta$ be the standard angle measured CCW from the positive $x$-axis. From trigonometry, if $z = r\hat{r} = x+iy$ then we can see that $x=r\cos \theta$ and $y= r \sin \theta$ hence apparently we must have $\hat{r} = (\cos \theta,\sin \theta)$ hence in our complex notation $\hat{r} = \cos \theta + i \sin \theta$.
What does $i^2=-1$ have to do with anything?
Let us suppose $i^2=-1$ and we multiply complex numbers as you were (hopefully) taught in an earlier part of your education,
$$ (a+ib)(c+id) = ac-bd+i(ad+bc) $$
Consider what happens when we multiply the direction vectors of two complex numbers with standard angle $\theta_1, \theta_2$ respective:
\begin{align} \hat{r}_1\hat{r}_2 &=(\cos \theta_1+i\sin \theta_2)(\cos \theta_2+i\sin \theta_2) \\
&= \cos \theta_1\cos \theta_2-\sin \theta_1\sin \theta_2+i \left[
\sin \theta_1\cos \theta_2+\sin \theta_2\cos \theta_1 \right] \\
&= \cos( \theta_1+\theta_2)+i \sin(\theta_1+\theta_2)
\end{align}
where we used the adding angles formulas in the last step (which you can have them derive from the law of cosines and the appropriate triangle). We find $i^2=-1$ paired with the usual arithmetic implies the product of unit-vectors in the complex plane gives us a new unit-vector in the direction which has standard angle $\theta_1+\theta_2$. If we use the notation $\hat{r}_1=e^{i\theta_1}$ and $\hat{r}_2=e^{i\theta_2}$ then we have shown that
$$ e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1+\theta_2)} $$
Of course, you might have assumed this anyway, but we have proved this law as a consequence of trigonometry you already know and the rule $i^2=-1$. In a nutshell, what makes the complex plane with points $z=x+iy$ the genuine complex plane is precisely the rule $i^2=-1$. Most everything else is calculus and algebra you already know.
From a different perspective, if you accept $e^{i\theta} = \cos \theta + i \sin \theta$ by faith (which is not without reason given the preceding paragraphs) this is a terribly useful way to recall trig. identities. It is easy to derive
$$ \cos \theta = \frac{1}{2}\left( e^{i\theta}+e^{-i\theta} \right)
\qquad \& \qquad \sin \theta =\frac{1}{2i}\left( e^{i\theta}+e^{-i\theta} \right) $$
Hence,
\begin{align}
\cos A \cos B &= \frac{1}{2}\left( e^{iA}+e^{-iA} \right)\frac{1}{2}\left( e^{iB}+e^{-iB} \right) \\
&= \frac{1}{4}\left( e^{i(A+B)}+e^{-i(A+B)} +e^{i(A-B)}+e^{-i(A-B)} \right)\\
&= \frac{1}{2}\left( \cos(A+B)+\cos(A-B)\right).
\end{align}
Could you derive the identity above before? Perhaps you were taught how to derive it via trigonometry, notice we just derived is from a logically linear argument using mere algebra. In fact, with Euler's Identity $e^{i\theta}= \cos \theta+i\sin \theta$ in hand we can derive nearly any trigonometric identity in real notation you desire. This is the basis for the success of the phasor methor, we are able to put into place some subtle trigonometric relations via relatively simple algebraic calculations with these so-called imaginary exponentials.
I try to introduce these ideas as soon as possible if I'm teaching a course where serious complex arithmetic is utilized. Notice, I fail to even distinguish between polar and exponential form in this answer. I'm sorry about that, they seem to be the same thing. Much like the abuse $x+iy=(x,y)$.
Finally, it's fun to think some about $j$, well, fine, $k$ such that $k^2=1$. In that plane with algebra $(a+bk)(c+dk)=ac+bd+k(ad+bc)$ we would find a very different story. It happens $e^{k\phi}= \cosh \phi + k \sinh \phi$, but I don't motivate that geometrically. Rather, analytical arguments force those hyperbolic sines and cosine to arise from the exponential series. Moreover, there are numbers with no multiplicative inverses along the numbers $z=r(1\pm k)$ for $r \in \mathbb R$. The analog of exponential representation breaks into four distinct cases in the regions separated by $y= \pm x$. In fact, there is some connection with Special Relativity which can be made here with these so-called hyperbolic numbers. For example, there is a paper (Hyperbolic Calculus by Motter and Rosa, Advances in Applied Clifford Algebras 8 No. 1, 109-128 (1998)) which solves the wave equation on the uniformly accelerated rigid rod by a clever application of such numbers. Personally, knowing what happens when we study non-complex numbers has given me much greater appreciation for $i^2=-1$.
All this said, the fundamental theorem of algebra is reason enough to study $\mathbb C$.
