There is an elementary proof which considers the polynomial $ p(z) $, $z \in \mathbb{C}$ as a function of $ (r,\theta) $ where $ z= r e^{-i \theta} $, $r\ge 0$ . The only assumption is that $z^{n}=a $, $a \in \mathbb{R}$ has a solution for all integer $n$. The only elementary theory used are as follows-

1. Properties of continuous functions on two dimensional plane. One important point is that continuous functions with real values divide the plane into two regions, positive and negative. The zeros separating these regions form a continuous curve or region.

2. Real and complex part of a continuous complex function are continuous.

3. The value of a polynomial $p(x) , x \in \mathbb{R}$ with real coefficients is dominated by its higher power for large absolute values of $x$.

4. For a continuous function with real values on the complex plane, intermediate value theorem holds on any continuous curve on the complex plane. This can be proved by parametrization of the curve.

5. $Re(z) < |z|, \forall z \in \mathbb{C}$.

The idea is to track the real and imaginary parts of a polynomial $p(r,\theta)$ separately. First showing that the real part $ Re(p(r,\theta))$ has atleast one zero on a big circle $r=R$ on the complex plane, we can further show that the curve containing this zero is open. Further analysis shows that the values of imaginary part of $p(r,\theta), Im(p(r,\theta))$ is positive on some point of this curve and negative at another for large $r$. Then $Im(p(r,\theta))=0$ somewhere on the particular curve $Re(p(r,\theta))=0$ we were tracking. This gives a solution for $p(r,\theta)=0$.

The proof can be visualized in the following way-

Writing $ z= r e^{-i \theta} $ we can find a big enough $r=R$ such that $ Re(p(R,\theta)) $ is positive at $\theta=\frac {2\pi k} {n} + \frac {\phi} {n} $ and negative at $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} $. This is because $Re(p(R,\theta)) $ is dominated by $r^{n}cos(n\theta + \phi)$. $\phi$ is the angle of the complex coefficient of $z^{n}$. So $Re(p(R,\theta))=0$ at some point in the interval $\theta= \left( \frac {2\pi k} {n} + \frac {\phi} {n} , \frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} \right) \forall k \in \mathbb{Z}$. So $ Re(p(r,\theta))=0 $ is nonempty. As a consequence of this and the continuity of $ Re(p(r,\theta)) $, $ Re(p(r,\theta))=0 $ containing this point forms a continuous curve on the complex plane.

The arc $r=R$ between $\theta=\frac {2\pi k} {n} + \frac {\phi} {n}$ and $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} $ is called a even sector for even $k$ and is called a odd sector for odd $k$. On these sectors, the leading coefficient of $Im(p(r,\theta))$ tends towards $\infty$ in the even sectors and towards $-\infty$ in the odd sectors. So if any of the continuous curves $Re(p(r,\theta))=0$ cross the arc $r=R$ in an even sector and an odd sector, we must have a root on this curve.

If we increase $r \ge R$ continuously on any sector, we get a zero for $Re(p(r,\theta))=0$ for some $\theta$. Continuity of $p(r,\theta)$ implies that we get a continuous curve for $ Re(p(r,\theta))=0 $ in any sector for $r\ge R$. Now, zeros of a continuous function divide the plane into two disconnected parts. One of the parts is finite in area if the curve is closed. Otherwise both areas are unbounded. Here, since $Re(p(r,\theta) \ne 0$ $ \forall r \ge R$ and $\theta=\frac {2\pi k} {n} + \frac {\phi} {n} $ or $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n} $, $Re(p(r,\theta))=0$ cannot be closed.

Now, there is some curve $Re(p(r,\theta))=0$ which crosses the arc $r=R$ odd number of times on any sector. If we start from an even sector, it must cross $r=R$ on another sector which is odd. This can be demonstrated using exhaustion since there are only finite numbers of arcs (a sequence of $2m$ consecutive integers cannot be paired in only disjoint or nested (even,even) and (odd,odd) pairs). Let us call this curve $f(r,\theta)$.

$Im(p(R,\theta)) $ is dominated by $r^{n}sin(n\theta + \phi)$. So, on $f(r,\theta)$, $Im(p(r,\theta)) $ is positive somewhere in the even sector and negative somewhere in the odd sector for big enough $r$. So $Im(p(z)=0) $ somewhere on the continuous curve $f(r,\theta)$. This gives a solution for $p(r,\theta)=0$.

Additionally, $f(r,\theta)$ is not asymptotic to $\theta=\frac {2\pi k} {n} + \frac {\phi} {n}$ or $\theta=\frac {2\pi k} {n} + \frac {\pi} {n} + \frac {\phi} {n}$. This can be proven by using the fact that $Re(z) < |z|, \forall z \in \mathbb{C}$. If this was not true, domination of $Im(p(r,\theta))$ on $f(r,\theta)$ by $r^{n}$ would be problematic since $sin(n \theta + \phi)$ might tend towards zero for increasing $r$.

Finally, for every root found, we can do division algorithm to get another polynomial of degree $n-1$. We know that quartic and lower order polynomials with complex roots are always solvable using radicals. So induction shows that all polynomials with real coefficients are solvable on the complex plane.