<s>[INST] A number theory exercise involving the floor function The problem is to prove that for integers $a,b$ with $b > 0$
$$\left\lfloor \frac{1}{b} \left\lfloor \frac{a}{b} \right\rfloor \right\rfloor = \left\lfloor \frac{a}{b^2} \right\rfloor.$$
I did this by obtaining an inequality both ways. First
$$k = \left\lfloor \frac{1}{b} \left\lfloor \frac{a}{b} \right\rfloor \right\rfloor \leq  \frac{1}{b} \left\lfloor \frac{a}{b} \right\rfloor \leq \frac{a}{b^2}. $$
Since $k$ is an integer,$$\left\lfloor \frac{1}{b} \left\lfloor \frac{a}{b} \right\rfloor \right\rfloor \leq \left\lfloor \frac{a}{b^2} \right\rfloor.$$
Then
$$\left\lfloor \frac{a}{b^2} \right\rfloor = \left\lfloor \frac{nb^2 +r}{b^2} \right\rfloor,$$
with $0 \leq r < b^2$ an integer.
$$k' = \left\lfloor \frac{nb^2 +r}{b^2} \right\rfloor = n = \frac{1}{b} \left\lfloor nb \right\rfloor \leq \frac{1}{b} \left\lfloor nb + \frac{r}{b} \right\rfloor = \frac{1}{b} \left\lfloor \frac{a}{b} \right\rfloor.$$
Since $k'$ is an integer
$$\left\lfloor \frac{a}{b^2} \right\rfloor  \leq \left\lfloor \frac{1}{b} \left\lfloor \frac{a}{b} \right\rfloor \right\rfloor $$ and we are done.
This is the first problem of the course and considering that, this solution feels insanely difficult. Is there an easier way to do this?
 [/INST] 
Is there an easier way to do this?

How about the following way?
There exist integers $m,r$ such that 
$$a=bm+r\quad\text{and}\quad 0\le r\lt b$$
from which
$$bm\le bm+r\lt b+bm\implies \frac mb\le \frac{bm+r}{b^2}\lt \frac{m+1}{b}$$
follows.
Since there is no integer $n$ such that
$$\frac mb\lt n\lt \frac{m+1}{b}$$
we get
$$\left\lfloor\frac{a}{b^2}\right\rfloor=\left\lfloor\frac{bm+r}{b^2}\right\rfloor=\left\lfloor\frac mb\right\rfloor=\left\lfloor\frac 1b\left\lfloor\frac ab\right\rfloor\right\rfloor$$
</s>