How to prove this identity of ceiling function? My book writes down this identity of least integer function:
$$\lceil x\rceil +\left\lceil x + \frac{1}{n}\right \rceil + \left\lceil x + \frac{2}{n}\right \rceil + \cdots +\left\lceil x + \frac{n -1}{n}\right \rceil = \lceil nx\rceil + n-1 $$. 
It didn't deduce it, however. I googled a bit about ceiling function but couldn't find any deduction. It is more like Hermite's Identity of floor function. Can anyone show me how to deduce this?

Another way of doing it is $$f(x)=\lceil x\rceil+\left\lceil x+\frac1n\right\rceil+\dots +\left\lceil x+1-\frac1n\right\rceil  - \left\lceil nx\right\rceil$$
We have $$
\begin{align}
f\left(x+\frac{1}{n}\right)=&\left\lceil x+\frac1n\right\rceil+\left\lceil x+\frac2n\right\rceil+\dots +\left\lceil x+1\right\rceil  - \left\lceil nx+1\right\rceil \\
=&\left\lceil x+\frac{1}{n}\right\rceil+\left\lceil x+\frac2n\right\rceil+\dots +\left\lceil x\right\rceil  - \left\lceil nx\right\rceil \\
=&\lceil x\rceil+\left\lceil x+\frac1n\right\rceil+\dots +\left\lceil x+1-\frac{1}{n}\right\rceil  - \left\lceil nx\right\rceil\\
=&f(x)
\end{align}$$
So $f$ is $\frac1n$-periodic. And we only need to evaluate it over $\left(0,\frac1n\right]$, in which case
$$f(x) = n-1$$
This is because if $x\in(0,\frac{1}{n}]$, then $x,x+\frac{1}{n},\dots,x+\frac{n-1}{n}$ and $nx\in(0,1]$ and $$\left\lceil x\right\rceil = \left\lceil x+\frac{1}{n}\right\rceil=\dots=\left\lceil x+\frac{n-1}{n}\right\rceil=\left\lceil nx\right\rceil=1$$
