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?