Why is renormalization (instead of regularization) is needed in QFT? This question looks like a duplicate question but I will try to make it different.
In QFT, since divergence arise in the calculation of some quantities, we need regularization to remove the infinities.
Now here comes the question: Why do we need renormalization then? I understand that renormalization can let us understand the theory in different energy scales. In statistical mechanics, it can lead to a better understanding of the system in different scales. But in QFT, isn't there only one scale (the energy scale our measurements can reach)? It seems to me that renormalization is a useful but not necessary skill. Then why do we need it in QFT?
 A: Because you can't just regularize a theory and say that that's it. You need to show that


*

*You are actually free to introduce a regularization and you didn't just come up with it for the sake of obtaining an answer.

*The choice of regularization scheme and of the cutoff does not affect physical quantities.

*You can in principle take a limit in which the regulator disappears and the resulting theory correctly represents a UV complete description of your particular model.


This is from the point of view of UV complete theories, where renormalization is all about taking the limit in which the energy cutoff is sent to infinity. Ok, you might say that one can use dimensional regularization, where there seems to be no cutoff. But then again, that's a trick and you need to argue that taking $d = 4 - \epsilon$ is a sensible thing to do. You can argue for that only using renormalization.
But if your viewpoint is to say that we only use QFT at a certain fixed scale $\Lambda_{\mathrm{LHC}}$, then ok, you can keep that scale as your cutoff. But then you measure some physical mass or whatever and you get a number 
$$
m^2 = f(\Lambda_{\mathrm{LHC}}, \mathrm{couplings},\ldots) = \mathrm{(some\; number)\;GeV}\,.
$$
And this is the point where you have to renormalize. In the sense that you fix the values of your couplings and whatnot in order to make the above relation hold.
