Exponential convergence of p-method $u'' + u = 2 \sin(x)$

Question

Consider the problem \begin{equation} -u'' + u = 2 \sin(x) \mbox{ in } (0, \pi), \hspace{2mm} u(0) = 0, u'(\pi) = -1. \end{equation} I want to show that a solution of the problem satisifies the following: For every $b > 0$ there exists $K_b > 0$ such that \begin{equation} \inf_{v \in \mathcal{P}_n}(||u-v||_{L^{\infty}(0,\pi)} + ||(u-v)'||_{L^{\infty}(0,\pi)}) \leq K_b e^{-bn} \mbox{ for all } n \in \mathbb{N}_0. \end{equation} where $\mathcal{P}_n$ is the space of all polynomials with coefficients in $\mathbb{C}$ of degree $= n$. The solution is $u(x) = \sin(x) \in C^{\infty}(0,\pi)$. For the approximation I use the approximation at chebyshev nodes by chebyshev polynomials \begin{equation} |u(x) - p_{n-1}(x)| \leq \frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n \max_{\xi \in (0,\pi)}|u^{(n+1)}(\xi)| \leq \frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n . \end{equation} Therefore \begin{equation} ||u - p_{n-1}||_{L^{\infty}} \leq \frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n . \end{equation} I stuck at estimating $\frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n$ to get the result above for some given $b > 0$. I tried using the stirling formula for $n!$ but this didn't work for me.