Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$ 
Let $x \in \Bbb R$ and $n \in \Bbb Z$ with $n \gt 0$. Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$;
in particular, $\left\lfloor\dfrac{\left\lfloor\dfrac {a}{b} \right\rfloor}{c} \right\rfloor = \left\lfloor\dfrac{a}{bc}\right\rfloor$

I don't know how to approach this problem, any hints to solve this would be greatly appreciated.
Thanks for your help!

Let $m=\lfloor x\rfloor$. Then $m\leq x < m+1$. So $\dfrac{m}{n}\leq \dfrac{x}{n}< \dfrac{m+1}{n}$. By the division algorithm, $m =qn+r$ for some integers $q,r$ with $0\leq r\leq n-1$. Thus, $$q\leq\dfrac{m}{n}\leq\dfrac{x}{n}<\dfrac{m+1}{n}=\dfrac{qn+r+1}{n}\leq q+1.$$
So $\lfloor \tfrac{m}{n} \rfloor=\lfloor\tfrac{x}{n}\rfloor=q$.
Note: $\lfloor x\rfloor$ is the unique integer $N$ such that $N\leq x<N+1$.
