Prove that, $\left\lfloor{\frac{x}{n}}\right\rfloor=\left\lfloor{\frac{\lfloor{x}\rfloor}{n}}\right\rfloor$ where $n \in{\mathbb{N}}$ 
Prove that
$$\left\lfloor{\frac{x}{n}}\right\rfloor=\left\lfloor{\frac{\lfloor{x}\rfloor}{n}}\right\rfloor,$$ where $n \in{\mathbb{N}}.$

My Attempt:
Let $x=nt$.
Then, I need to prove,
$$\lfloor{t}\rfloor=\left\lfloor{\frac{\lfloor{nt}\rfloor}
{n}}\right\rfloor.$$
Let $t=n\lambda+r+f$, 
where $0\leq{r}\leq{n-1}$ and $0\leq{f}\lt{1}.$
This gives R.H.S as,
$$\left\lfloor{\frac{\lfloor{n^2\lambda+nr+nf}\rfloor}
{n}}\right\rfloor.$$
How do I continue?
I also know the property,
$$\lfloor{x}\rfloor=\left\lfloor{\frac{x}{n}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{n}}\right\rfloor+\left\lfloor{\frac{x+2}{n}}\right\rfloor \cdots \left\lfloor{\frac{x+(n-1)}{n}}\right\rfloor.$$
Can I use that here?

It is actually much simpler.
See, $\left\lfloor \frac xn\right\rfloor$ is the unique integer $l$ such that $l \leq \frac xn \leq l+1$, or in other words, $nl \leq x \leq nl+n$.
Note that $nl$ is an integer less than or equal to $x$, and $\lfloor x \rfloor$ is the greatest such integer, so $\lfloor x \rfloor \geq nl$. Of course, $\lfloor x \rfloor \leq x \leq nl + n$. 
Consequently, $nl \leq \lfloor x \rfloor \leq nl + n$. Dividing by $n$, one sees the conclusion.
