Using the power series expansion $\tan(x) = \sum_{n=0}^{+\infty} \frac{\alpha_{2n+1}}{(2n+1)!} x^{2n+1}$ and the formula $\pi \tan(\pi x) = \sum_{p=1}^{+\infty} 2(2^{2p}-1)\zeta(2p) x^{2p-1}$, show
$$\forall n \in \mathbb{N}, \quad \alpha_{2n+1} = \frac{2\left(2^{2n+2} - 1\right)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2).$$