How to prove $\left \lceil \frac{n}{m} \right \rceil = \left \lfloor \frac{n+m-1}{m} \right \rfloor$?

If we do a little algebra on the right-hand side, we get:

$$\left\lfloor \frac{n+m-1}{m} \right\rfloor = \left\lfloor \frac{n-1}{m} \right\rfloor + 1$$

Hopefully things should be clear from that point on. :)

A vivid conceptual proof: $\:$ If $\rm\:[a,b]\:$ contains a unique integer $\rm\:k\:$ then clearly $\rm\ \lceil a\rceil\: =\: k\: =\: \lfloor b\rfloor\:.$

This applies to $\rm\:\ a = n/m,\:\ b = (n+m-1)/m\:.\:$ Notice that $\rm\:[a,b]\:$ contains a unique integer since $\rm\:m\:$ divides exactly one of the consecutive $\rm\:m\:$ integers $\rm\:n,\:n+1,\:\cdots,\:n+m-1\:.$

Note $\ $ This problem is exercise 3.12 in Graham; Knuth; Patashnik: Concrete Mathematics. Curiously they overlook this simple solution, instead giving essentially the solution in my other answer here.

HINT

We can always write $n = mk - r$ where $0 \leq r < m$ and hence $$\left \lceil \frac{n}{m} \right \rceil = k.$$ Try to argue from this what $\left \lfloor \frac{n+m-1}{m} \right \rfloor$ is.

By the division algorithm $\rm\ n\: =\: k\ m + r,\ \ 0\le r < m\:.\:$ Substituting, the sought identity becomes

$$\rm \left\lceil{\ k\: +\: \frac{r}m}\ \right\rceil\ =\ \left\lfloor{\ k\: +\: \frac{r+m-1}m\ }\right\rfloor$$

This is true since both sides are equal to $\rm\:k\:$ if $\rm\:r = 0\:,\:$ else equal to $\rm\:k+1\ $ if $\rm\ 1 \le r \le m-1\:$