Proving by induction that $frac{n(n + 1)(2n + 1)}{6} = 0^2 + 1^2 + 2^2 + 3^2 + ... + n^2$

Note: I am asking this question as a simple introductory question to proofs by induction, to which I will give also my formal answer (which should be correct, if not, please comment) for future visitors of this website.

I know that to prove something by induction, we need formally 3 steps (2, if we considere inductive hypothesis and inductive step together):

1. Base case or basis:

   
   
   

   We determine what is the base case for our statement $P(n)$, and we immediately prove it.




2. Inductive hypothesis:

   
   
   

   We assume our statement $P(n)$ is true for $n$.




3. Inductive step:

   
   
   

   We try to prove that our statement is also true for $P(n + 1)$. If we can prove that, we can also prove that our statement is true for $n + 2$ (we don't need to do it), and so on.

   
   
   
   

   This step is usually the most difficult part, because it involves some algebraic manipulation, or some imagination in some way. For this last reason, it might be helpful watching other similar proofs by induction to solve the current one.

For example, suppose our $P(n)$ is the following:

$$\frac{n(n + 1)(2n + 1)}{6} = 0^2 + 1^2 + 2^2 + 3^2 + ... + n^2$$

which is basically saying that the sum of all squares of the natural numbers from $0$ to $n$ can be determined by the formula on the left side of the equation.

How would we prove $P(n)$ is true for all $n \in \mathbb{N}$?

For more information about the proof by induction, see the Wikipedia page about induction called Mathematical induction.