Why is it considered that $(\mathrm d x)^2=0$? Why is it okay to consider that $(\mathrm d x)^n=0$ for any n greater than $1$?
I can understand that $\mathrm d x$ is infinitesimally small ( but greater than $0$ ) and hence its square or cube should be approximately equal to $0$ not exactly $0$ . 

But if this is so then how can we expect the results obtained from calculus to be exact and not just approximate ( like the slope or area under a curve )? 

I have also noticed some anomalies, like $\sqrt{ (\mathrm d x)^2 + (\mathrm d y)^2 }$ is $0$ but $\mathrm d x\sqrt{1+ (\mathrm d y/\mathrm d x)^2 }$ is not $0$ when these two things are apparently the same . Moreover we can  claim that 

$$(\mathrm d x)^2=(\mathrm d x)^3=(\mathrm d x)^4 = \cdots = 0$$

which is quite hard to believe.

Can you help me figure out the logic behind these things ?
It is not the square of $\mathrm{d}x$ that is $0$. It is $\mathrm{d}x\land\mathrm{d}x$ that is zero. 

This comes into play in differential form and in changes of variables. Suppose that $u=x+y$ and $v=x-y$. Then
$$
\begin{align}
\iint f\,\mathrm{d}u\,\mathrm{d}v
&amp;=\iint f\,\mathrm{d}(x+y)\,\mathrm{d}(x-y)\\
&amp;=\iint f\,\mathrm{d}x\,\mathrm{d}x+\iint f\,\mathrm{d}x\,\mathrm{d}y-\iint f\,\mathrm{d}y\,\mathrm{d}x-\iint f\,\mathrm{d}y\,\mathrm{d}y\\
&amp;=\iint f\,\mathrm{d}x\,\mathrm{d}y-\iint f\,\mathrm{d}y\,\mathrm{d}x\tag{1}
\end{align}
$$
Why do $\iint f\,\mathrm{d}x\,\mathrm{d}x=\iint f\,\mathrm{d}y\,\mathrm{d}y=0$? Well, inside the outer integral, $x$ is supposed to be held constant, so the inner $\mathrm{d}x$ will vanish. The same goes for the double $y$ integral.

Another consequence of this follows from
$$
\begin{align}
0
&amp;=\iint f\,\mathrm{d}u\,\mathrm{d}u\\
&amp;=\iint f\,\mathrm{d}(x+y)\,\mathrm{d}(x+y)\\
&amp;=\iint f\,\mathrm{d}x\,\mathrm{d}x+\iint f\,\mathrm{d}x\,\mathrm{d}y+\iint f\,\mathrm{d}y\,\mathrm{d}x+\iint f\,\mathrm{d}y\,\mathrm{d}y\\
&amp;=\iint f\,\mathrm{d}x\,\mathrm{d}y+\iint f\,\mathrm{d}y\,\mathrm{d}x\tag{2}
\end{align}
$$
That is, $\mathrm{d}y\land\mathrm{d}x=-\mathrm{d}x\land\mathrm{d}y$. Thus, the integral in $(1)$ is equal to
$$
2\iint f\,\mathrm{d}x\,\mathrm{d}y\tag{3}
$$



Note

It is not the case that $\sqrt{\mathrm{d}x^2+\mathrm{d}y^2}=0$. It is the same as $\mathrm{d}x\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}$ in whatever context they both make sense.
A little historical perspective: When mathematicians back in the 1500's - 1700's were solving problems using 'infinitesimals' - the backdrop of the invention of Calculus - their arguments were exactly what you are complaining about.  And people then complained about it too:

If 'h' isn't zero, you have no right to just drop the term as if it were; if 'h' is zero, you have no right to divide by it.  And don't give me that 'infinitesimal' nonsense, it has to be zero or nonzero.

To which the general reply was something on the order of:

Well, it works, so: get out of my face.

And the amazing thing was, it really did work.  But that was mostly because the early practitioners were basically geniuses - Newton, Leibniz, Bernoulli, Cauchy, Euler, etc.  It was only about a hundred years later, when people started trying to teach Calculus to us mere mortals, that the old 'hand-waving' arguments wouldn't do any more.  That's when you get Bolzano, Weirstrauss, etc., creating the notion of 'limit' and epsilon-delta arguments to make it all logically robust, partly so their students would 'buy' their arguments in class.  [that's not the only reason for the clean-up, but it was an important one]