How to Show that $1+\left(\left\lceil\dfrac{x}{n}\right\rceil -1\right)n\leq x$? For $x$ integer in $\{1,\ldots,n^2\}$. I would like to show that 
$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n \leq x.
$$
I start by the property of the ceiling function which gives me
$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n &lt;x+1.
$$
By the property of the ceiling function we have:

$$
\left\lceil\dfrac{x}{n}\right\rceil &lt;\dfrac{x}{n}+1,
$$
which gives me:
$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n &lt;x+1.
$$

What I missed which seems some kind of obvious is that $\left(1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n\right)$ is a positive integer and since it is less than $x+1$, it must be less than or equal to $x$. Thus,

$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n \le x.
$$
Okay divide both sides by $n$, then:
$$\left\lceil\dfrac{x}{n}\right\rceil-1 \le \frac{x-1}{n}$$ then we got:
$$\left\lceil\dfrac{x}{n}\right\rceil = \frac{x+k}{n}$$ where $k$ some integer number.
So we have inequality with $k$ and $n$ :
$$n \ge k-1$$ , now think about what set of valid $k$ and you will prove it.