Real Analysis. Differentiation (Taylor's Formula). Check proof. 
Show that the polynomial
  $$x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... \pm \frac{x^n}{n!},$$
  with $n$ odd, differs from $\sin x$ by, at most, $\frac{\pi^{n+1}}{(n+1)!}$ in $[-\pi, \pi]$.

$\textbf{Solution:}$ Applying the Taylor's Formula in $f(x) = \sin x$ for $x_{0} = 0$:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... \pm \frac{x^n}{n!} + R_{n+1}$$
and
$$R_{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$
for $c \in (0,x)$. Since $|f^{(n+1)}(c)| \leq 1$ and $|x|^{n+1} \leq \pi^{n+1}$
$$R_{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \leq \frac{\pi^{n+1}}{(n+1)!}.$$
The result follows.
Is this correct?

Yes, your proof is correct.
I would change $$ R_{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \leq \frac{\pi^{n+1}}{(n+1)!}.$$ 
to $$ |R_{n+1}| = \frac{|f^{(n+1)}(c)|}{(n+1)!}x^{n+1} \leq \frac{\pi^{n+1}}{(n+1)!}.$$ 
