When $n = 1$, it obviously holds that $|\sin x| \le |\sin x|$. Assuming that $n=k$, the relation still holds, i.e. $|\sin (kx)|\le k|\sin x|$. Then, when $n=k+1$: \begin{align} &|\sin((k+1)x)|\\ &=|\sin(kx)\cos{x} + \cos(kx)\sin{x}|\\ &\le |\sin(kx)\cos{x}| + |\cos(kx)\sin{x}|\\ & \le |sin(kx)| + |sin(x)| \\ & < k|\sin{x}| + |\sin{x}|\\ &=(k+1)|\sin{x}| \end{align} Therefore, by the principle of mathematical induction, the relation holds for all positive integer $n$.