Prove $\sin(1/n)<1/n$ for all $n$

I need to prove $\sin(1/n)<1/n$ for all $n \in \Bbb N$ using mathematical induction.

Dont know how to start. Please help!

Because of \[ \sin(x)=x-\frac{x^3}{3!}+ \frac{x^5}{5!}-\frac{x^7}{7!} +\dots \] As $x^n$ is monotone decreasing on $[0,1]$ in respect to $n$. Hence \[ \sin(x)\leq x \] for $x$ in $[0,1]$. As $\sin(x)\leq 1$ the inequality is even true for all $x\in [0,\infty)$.

We can see that when $x-\sin(x)> 0$ holds for $x\in (0,1]$ this implies that $\sin\big(\frac{1}{n}\big) < \frac{1}{n}$ holds for all $n \in \mathbb{N}$. At first we note that for $x=0$ $\sin(x)-x=0$. The derivative of \[ x - \sin(x)\] is $1-\cos(x)$. With the fundamental theoroem of calculus we know that \[ x-\sin(x) = (1-\cos(\xi)) \cdot x \] where $\xi \in (0,x)$. As $\cos (x) \leq 1$ and $x>0$ we have that $x-\sin(x)$ is positive and hence our inequality holds.