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Here is an elementaryproof which considers $ p(z) $ as a function of $ (x,y) $ where $ z=x+iy $.

The only assumptions are that $z^{n}=r $ has a solution for all integer $n$ and real $r$ and that intermediate value theorem holds for continuous real valued functions along any continuous curve on the complex plane. The second assumption can be proved by parametrizing the curve and then using the least upper bound principle for real numbers.

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 $r^{n}cos(n\theta)$. $\phi$ is the angle of the complex coefficient of $z^{n}$.

Now , $ p(z) $ being continous, $ Re(p(z)) $ is also continuous. So the zeros of $ Re(p(z)) $ form a continuous curves on the complex plane. The point is to show that on one such continuous curve there is a point where $ Im(p(z))>0 $ and another point where $ Im(p(z))<0 $. Then on the particular curve $ Re(p(z))=0 $ there must be a point where $ Im(p(z))=0 $. However, this is the solution of $ p(z)=0 $.

The proof can be visualized in the following way-

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 $k$.

Let us start from the even arc with $k=0$. The values of $ Re(p(R,\theta)) $ at the ends of this arc are of opposite sign, so it much be zero somewhere on the curve. The same thing holds if we increase $r \ge R$ continuously. So we get a continuous curve for $ Re(p(z))=0 $ in the sector $\theta \in$ { $0 $, $\frac {\pi} {n}$} for $r\ge R$. Let us call this curve $f(z)$ 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 we cannot have $ Re(p(z))=0 $ for $\theta =0 $ and $\theta=\frac {\pi} {n}$ for $r\ge R$, the curve $f(z)$ does not intersect the rays $\theta =0 $ and $\theta=\frac {\pi} {n}$ for $r \ge R$. This means $f(z)$ cannot be closed.

Now, $f(z)$ must cross one of the odd arcs. This can be constructed using exhaustion (there are only finite numbers of arcs). If not, there must be another curve satisfying $ Re(p(z))=0 $ and the desired conditions.

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

Additionally, $f(z)$ 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 $Im(p(z))$ on $f(z)$ by $r^{n}$ would be problematic since $sin(n \theta)$ might tend towards zero.

Finally, for real coefficients, if $z$ a solution, $\overline z$ is also a solution for $p(z)=0$, since $\overline p(z) = p(\overline z)=0$. So for every root found, we can do division algorithm to get another polynomial of degree $n-2$. We know that quartic polynomials are always solvable using radicals. So induction shows that all polynomials with real coefficients are solvable on the complex plane.