If it's a slick proof you want, nothing beats this proof that the only cases where both $r$ and $\cos(r \pi)$ are rational are where $\cos(r \pi)$ is $-1$, $-1/2$, $0$, $1/2$ or $1$.
If $r=m/n$ is rational, $e^{i \pi r}$ and $e^{-i \pi r}$ are roots of $z^{2n} - 1$, so they are algebraic integers. Therefore $2 \cos(r \pi) = e^{i \pi r} + e^{-i \pi r}$ is an algebraic integer. But the only algebraic integers that are rational numbers are the ordinary integers. So $2 \cos(r \pi)$ must be an integer, and of course the only integers in the interval $[-2,2]$ are $-2,-1,0,1,2$.
I considered the more general formula :
$$T(m):=\prod_{k=1}^m \tan\left(\frac{k\pi}{4m}\right)$$
and noticed that the result for small values of $m$ was solution of a polynomial of degree $\le m$.
For $m=45$ I found that the answer was solution of this irreducible polynomial of degree $24$ :
$$1\\- 3256701697315828896312\,x^1 - 325994294876282580655116\,x^2 + 7220097128841103979624568\,x^3 + 112578453555034444841119842\,x^4 + 493898299320136273975435032\,x^5 + 649061666980531722406164708\,x^6 - 840700351973464244018822232\,x^7 -2457988129238279755530778353\,x^8 - 138286882106888055215208624\,x^9 +2474525072938192662606171624\,x^{10}+326024084648343835216068912\,x^{11} - 1088043811994145989051965476\,x^{12} + 5147738954805237173669808\,x^{13} +182273284200850360076819304\,x^{14} - 33045263177263307887100976\,x^{15} + 677463542076505961377071\,x^{16} +170537100491574073221480\,x^{17} - 6714674580553776884700\,x^{18} - 128584156182235814952\,x^{19} - 339010000890501150\,x^{20}\\ +776030507612856\,x^{21} - 397610115660\,x^{22}\\ + 37004040\,x^{23} + 6561\,x^{24}$$
This is an 'experimental' result (no proof) but rather satisfying from the 'not nice' point of view ! ;-)
Best Answer
One cannot express $\tan(50^\circ)$ purely in terms of real radicals. For if one could, then one could express $\cos(20^\circ)$ in terms of real radicals, and it is known that one cannot do that. (It is an instance of the casus irreducibilis of the cubic.)
As is implicitly pointed out in the post, one can express $\tan(50^\circ)$ in terms of a primitive ninth root of unity.
Added: The edited question asks for the minimal polynomial of $\tan(50^\circ)$. Let $x=\tan(50^\circ)$. Using the identity $\tan(3\theta)=\frac{3\tan\theta -\tan^3\theta}{1-3\tan^2\theta}$, we find that $\frac{3x-x^3}{1-3x^2}=-\frac{1}{\sqrt{3}}$. Square and simplify. We get a sextic in $x$, which is irreducible by the Eisenstein Irreducibility Criterion.