We needs the following Lemma: If the polynomial $ P(x) = a_0 x^n +a_1 x^{n-1}+ ...+ a_{n-1}x + a_n \in Z[x], a_0 \ne 0$ admits as a root an irreducible fraction $ \frac {a}{b} \ne 0$, then $ a$ divides $ a_n$ and $ b$ divides $ a_0$. Now let us consider two cases. 1° $ m = 2k+1, k\in Z$, $ \tan(m \alpha )= \frac { \pm x ^{m} + P_m (x)}{1 + Q_m (x)}$ where $ x =\tan \alpha$, and $ P_m (x)$, $ Q_m (x)$ are polynomials of integer coefficients, of order at most $ m -1$. By induction, for $ m=1$ it is obvious. Let us suppose that is true for some $ m$, and consider $ \tan (m+2) \alpha =\tan ( m\alpha + 2 \alpha) = \frac{\tan m\alpha +\tan 2 \alpha}{1 - ( \tan m\alpha)( \tan 2 \alpha)} = (\frac{\pm x + P_m (x)}{1 + Q_m(x)} +\frac{2x}{1 - x^2}) / W(x)= \frac {\pm x^{m+2} \pm x^m + (1 - x^2) p_m (x) + 2x (1 + Q_m(x))}{1+ Q(x) - x^2(1+Q_m(x))-2x(\pm x\m + P_m(x))}$, (where $ W(x) = 1 - 2x ( \frac{ \pm x^m +P_m(x)}{(1-x^2) (1+Q_m(x))} )$ ), which imply that the representation is valid for $ m+2$. Now, let $ \alpha = \frac{ \pi}{m}$. It follows that $ \pm x_0^{m} + P_m (x_0) = 0$. The lemme implies that $ x_0 = tan \alpha$ must be integer. But we know, that $ 0 < \tan \alpha <1$, so contradiction. 2° $ m= 2^k \cdot n$, with $ n$ is odd. Since $ \tan (2 \alpha) = 2 ( \frac {\tan \alpha}{1- (\tan \alpha)^2})$, and supposing $ \ tan (\frac{ \pi}{ 2^k n} ) \in Q$ we have that all the terms of the following sequence $ \ tan (\frac{ \pi}{ 2^{k-1} n} )$, $ \tan (\frac { \pi}{ 2^{k-2} n} )$,.... $ \ tan (\frac{ \pi}{ 2^0 n} )$= $ \ tan (\frac{ \pi}{ n}$ ) are all rational. But that is contradiction. So that is done