Proving that: $\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}$

Since $$ \frac{\sin(x)}{x} = \int_{0}^{1} \cos(tx) dt$$

with proper justifications (differentiation under the integral sign) you may derive

$$\left(\frac{\sin(x)}{x}\right)^{(n)} = \int_{0}^{1} t^n \cos(tx+n\pi/2) dt$$ which immediately yields the wanted estimate.

See that, $\displaystyle\frac{\sin x}{x} =f(x) = \frac{1}{2}\int_{-1}^{1} e^{-itx} dt$ Then $$|f^{(n)}(x)| =\left|\frac{1}{2}\int_{-1}^{1} (-it)^ne^{-itx} dt\right| \le\frac{1}{2}\int_{-1}^{1} |t|^n dt=\int_0^1t^n\,dt=\frac1{n+1}.$$